Python Basic Exercise for Beginners: 40 Coding Problems with Solutions

Updated on: February 8, 2026 | 527 Comments

This Python exercise for beginners is designed to help you practice and improve your coding skills. This page contains over 40 Python exercises curated for beginners.

Each exercise includes a clear problem, a helpful hint, a complete solution, and a detailed explanation. This ensures you not only solve the problem but also understand why the solution works.

Also, See:

Python Exercises: A set of 17 topic specific exercises
Python Quizzes: Solve quizzes to test your knowledge of fundamental concepts.
Basic Python Quiz for beginners
Python Basics: Learn the basics of Python to solve this exercise.
Beginner Python Interview Questions

Tips and essential learning resources accompany each question. These will help you solve the exercise and become more familiar with Python basics.

This Python exercise covers questions on the following topics:

Python for loop and while loop
Python list, set, tuple, dictionary, input, and output
Use Online Python Code Editor to solve exercises.

Let us know if you have any alternative solutions in the comments section below. This will help other developers.

+ Table of Contents (40 Exercises)

Exercise 1. Arithmetic Product and Conditional Logic

Practice Problem: Write a Python function that accepts two integer numbers. If the product of the two numbers is less than or equal to 1000, return their product; otherwise, return their sum.

Exercise Purpose: Learn basic control flow and the use of if-else statements. Understand how code decisions change output based on a mathematical threshold.

Given Input:

Case 1: number1 = 20, number2 = 30
Case 2: number1 = 40, number2 = 30

Expected Output:

The result is 600
The result is 70

Refer:

Hint

Define a function that calculates the product two given numbers.
Use an if statement to compare that product against 1000 to decide whether to return the product or the sum.

Solution

def multiplication_or_sum(num1, num2):
    # Calculate product
    product = num1 * num2
    
    # Check if product is within the threshold
    if product <= 1000:
        return product
    else:
        return num1 + num2

# Testing Case 1
result = multiplication_or_sum(20, 30)
print("The result is", result)

# Testing Case 2
result = multiplication_or_sum(40, 30)
print("The result is", result)Code language: Python (python)

Explanation to Solution:

Function Definition: The code uses def to create a reusable block of logic, Inside the function multiply these two numbers and store their product in a variable.
Conditional Branching: The if product <= 1000 line acts as a gatekeeper, routing the program’s logic based on the calculated value.
Return Values: Instead of just printing, the function returns the value, allowing the calling code to decide how to display or use the result.

Exercise 2. Cumulative Sum of a Range

Practice Problem: Iterate through the first 10 numbers (0–9). In each iteration, print the current number, the previous number, and their sum.

Exercise Purpose: This exercise teaches “State Tracking.” In programming, you often need to remember a value from a previous loop iteration to calculate results in the current one. This is the basis for algorithms like Fibonacci sequences or running totals.

Given Input: Range: numbers = range(10)

Expected Output:

Printing current and previous number sum in a range(10)
Current Number 0 Previous Number 0 Sum: 0
Current Number 1 Previous Number 0 Sum: 1
Current Number 2 Previous Number 1 Sum: 3
....
Current Number 8 Previous Number 7 Sum: 15
Current Number 9 Previous Number 8 Sum: 17

Reference article for help:

Hint

Initialize a variable previous_num to 0 outside the loop.
At the end of each loop iteration, update previous_num with the value of the current_num.

Solution

print("Printing current and previous number sum in a range(10)")
previous_num = 0

# Loop from 0 to 9
for i in range(10):
    x_sum = previous_num + i
    print(f"Current Number {i} Previous Number {previous_num} Sum: {x_sum}")
    
    # Update previous_num for the next iteration
    previous_num = iCode language: Python (python)

Explanation to Solution:

Initialization: previous_num = 0 is set outside the loop so it persists across iterations.
for loop Iteration: The for i in range(10) loop automatically increments i from 0 to 9.
Next, display the current number (i), the previous number, and the addition of both numbers in each iteration of the loop.
State Update: The line previous_num = i is critical; it “shifts” the current value into the “memory” slot for the next cycle of the loop.

Exercise 3. String Indexing and Even Slicing

Practice Problem: Display only those characters which are present at an even index number in given string.

Exercise Purpose: Understand how data is stored in memory using zero-based indexing. In most languages, the first character is at position 0, the second at 1, and so on. Mastering indexing is vital for data parsing.

Given Input: String: "pynative"

Expected Output:

Original String is  pynative
Printing only even index chars
p
n
t
v

Hint

You can solve this using a for loop with a range that steps by 2. Iterate through the characters of the string using a loop and the range() function.
Use start = 0, stop = len(s) - 1, and step = 2. The step is 2 because we want only even index numbers.
Or by using Python’s built-in string slicing syntax [::2].

Solution

word = "pynative"
print("Original String is ", word)

# Method: Using list slicing
# format: [start:stop:step]
even_chars = word[0::2]

print("Printing only even index chars")
for char in even_chars:
    print(char)Code language: Python (python)

Explanation to Solution:

Zero-based Indexing: The character ‘p’ is at index 0, ‘y’ is at 1, ‘n’ is at 2.
Slicing [0::2]: This is a powerful Python feature. It tells the program: “Start at index 0, go to the end, and skip every 2nd character.”
Iteration: The for char in even_chars loop processes the resulting subset of characters individually.

Solution 2: Using loop

# accept input string from a user
word = input('Enter word ')
print("Original String:", word)

# get the length of a string
size = len(word)

# iterate a each character of a string
# start: 0 to start with first character
# stop: size-1 because index starts with 0
# step: 2 to get the characters present at even index like 0, 2, 4
print("Printing only even index chars")
for i in range(0, size - 1, 2):
    print("index[", i, "]", word[i])Code language: Python (python)

Exercise 4. String Slicing and Substring Removal

Practice Problem: Write a function to remove characters from a string starting from index 0 up to n and return a new string.

Exercise Purpose: This exercise demonstrates how to truncate data strings, a common data-cleaning task.

Given Input:

remove_chars("pynative", 4)
remove_chars("pynative", 2)

Expected Output:

tive
native

Hint

Use string slicing to get a substring
To remove the first n characters, you essentially want to keep everything from index n to the end of the string: string[n:].

Solution

def remove_chars(word, n):
    print('Original string:', word)
    # Extract string from index n to the end
    res = word[n:]
    return res

print("Removing characters from a string")
print(remove_chars("pynative", 4))
print(remove_chars("pynative", 2))Code language: Python (python)

Explanation to Solution:

Slicing word[n:]: By omitting the “stop” value in the slice, Python defaults to the very end of the string.
Flexibility: The function is designed to take n as an argument, making it reusable for any length of removal.
Memory: Note that strings in Python are immutable; this function doesn’t change the original string but creates and returns a new truncated version.

Also, try to solve Python string exercises

Exercise 5. Variable Swapping (The In-Place Method)

Practice Problem: Write a program to swap the values of two variables, a and b, without using a third temporary variable.

Exercise Purpose: This exercise will help you learn about memory efficiency and Python’s special tuple unpacking feature. In other languages like C or Java, you need a temporary variable to swap values safely. In Python, you can swap values in one line without risking data loss.

Given Input: a = 5, b = 10

Expected Output:

Before Swap: a = 5, b = 10
After Swap: a = 10, b = 5

Hint

In Python, you can use the syntax a, b = b, a.
This evaluates the right side (creating a tuple) and then assigns the values back to the left side simultaneously.

Solution

a = 5
b = 10
print(f"Before Swap: a = {a}, b = {b}")

# Simultaneous assignment (Tuple Unpacking)
a, b = b, a

print(f"After Swap: a = {a}, b = {b}")Code language: Python (python)

Explanation to Solution:

Simultaneous Evaluation: Python evaluates the entire right side (b, a) first, essentially holding both values in a temporary hidden structure (a tuple).
Assignment: It then unpacks those values into the variables on the left. This prevents the value of a from being overwritten by b before its original value can be moved.
Clean Code: This eliminates the need for three lines of code and an extra “temp” variable, making the script more readable.

Exercise 6. Calculating Factorial with a Loop

Practice Problem: Write a program that calculates the factorial of a given number (e.g., 5!) using a for loop.

Exercise Purpose: This exercise explores “Mathematical Accumulation.” A factorial (e.g., 5! = 5*4*3*2*1) requires you to maintain a running product across multiple iterations, which is a core pattern in scientific computing.

Given Input: number = 5

Expected Output: The factorial of 5 is 120

Hint

Initialize a factorial variable to 1.
Loop through a range from 1 to the given number, multiplying the factorial variable by the current loop index at each step.

Solution

num = 5
factorial = 1

# Loop from 1 to num (inclusive)
for i in range(1, num + 1):
    factorial = factorial * i

print(f"The factorial of {num} is {factorial}")Code language: Python (python)

Explanation to Solution:

Identity Element: We start factorial at 1 because multiplying by zero would ruin the entire calculation.
range(1, num + 1): Since the end of a range is exclusive, we use + 1 to ensure the loop includes the number 5.
Running Product: Each pass of the loop updates the value of factorial, building the final result step-by-step.

Exercise 7. List Manipulation: Add and Remove

Practice Problem: Create a list of 5 fruits. Add a new fruit to the end of the list, then remove the second fruit (at index 1).

Exercise Purpose: This exercise teaches “Dynamic Collection Management.” Lists are rarely static; being able to modify, expand, and prune them is essential for handling data like shopping carts, user lists, or inventory systems.

Given Input: fruits = ["apple", "banana", "cherry", "date", "elderberry"]

Expected Output: ['apple', 'cherry', 'date', 'elderberry', 'fig']

Hint

Use .append() to add an item to the end
and .pop(1) or del to remove the item at the specific index.

Solution

fruits = ["apple", "banana", "cherry", "date", "elderberry"]

# Add a fruit to the end
fruits.append("fig")

# Remove the second fruit (index 1)
fruits.pop(1)

print(fruits)Code language: Python (python)

Explanation to Solution:

.append(): This method adds the new item without disturbing the existing order.
Zero-Based Indexing: In Python, the “second” item is actually at index 1 (0 is the first).
.pop(1): This removes the item at index 1 and “shifts” the remaining items (cherry, date, etc.) to the left to fill the gap, maintaining list integrity.

Exercise 8. String Reversal

Practice Problem: Write a program that takes a string and reverses it (e.g., “Python” becomes “nohtyP”).

Exercise Purpose: This exercise demonstrates “Sequence Slicing.” Strings in Python are sequences, and mastering the slicing syntax is a powerful shortcut for data manipulation that would take 5-10 lines of code in other languages.

Given Input: text = "Python"

Expected Output: Reversed: nohtyP

Hint

Use Python’s slicing notation [start:stop:step].
Setting the step to -1 tells Python to move through the sequence backwards.

Solution

text = "Python"

# Reversing using slicing
reversed_text = text[::-1]

print(f"Original: {text}")
print(f"Reversed: {reversed_text}")Code language: Python (python)

Explanation to Solution:

[::-1]: The first two colons imply “start at the very beginning” and “go to the very end.” The -1 indicates the direction of travel.
Immutability: This operation doesn’t change the original text variable; instead, it creates a new string in memory with the characters in reverse order.
Efficiency: This is the most efficient way to reverse a string in Python, utilizing optimized internal C code.

Exercise 9. Vowel Frequency Counter

Practice Problem: Write a program to count the total number of vowels (a, e, i, o, u) present in a given sentence.

Exercise Purpose: This exercise introduces “Membership Testing.” By checking if a character belongs to a specific group (the vowels), you learn how to filter data based on categories. This is a fundamental step toward building text-analysis tools or spam filters.

Given Input: sentence = "Learning Python is fun!"

Expected Output: Number of vowels: 6

Hint

Define a string containing all vowels "aeiou".
Loop through the sentence, convert each character to lowercase to ensure the check is case-insensitive, and increment a counter if the character exists in your vowel string.

Solution

sentence = "Learning Python is fun!"
vowels = "aeiou"
count = 0

# Convert to lowercase to handle 'A' and 'a' equally
for char in sentence.lower():
    if char in vowels:
        count += 1

print(f"Number of vowels: {count}")Code language: Python (python)

Explanation to Solution:

.lower(): This is essential for robust code. It ensures that “A” and “a” are both counted without needing to write a massive if statement for both cases.
The in Keyword: This is a highly optimized Python operator that checks for the existence of an item within a sequence (the string of vowels).
Counter Pattern: We use a simple integer (count) that “accumulates” every time the condition evaluates to True.

Exercise 10. Finding Extremes (Min/Max) in a List

Practice Problem: Given a list of integers, find and print both the largest and the smallest numbers.

Exercise Purpose: This exercise explores “Aggregate Functions.” While Python has built-in tools for this, understanding how to identify extremes is critical for data normalization, where you often need to find the range of a dataset before processing it.

Given Input: nums = [45, 2, 89, 12, 7]

Expected Output: Largest: 89 Smallest: 2

Hint

Python provides the max() and min() functions which iterate through a collection and return the extreme values in O(n) time.

Solution

nums = [45, 2, 89, 12, 7]

largest = max(nums)
smallest = min(nums)

print(f"Largest: {largest}")
print(f"Smallest: {smallest}")Code language: Python (python)

Explanation to Solution:

max(nums): Internally, this function assumes the first element is the largest, then compares it against every other element in the list, updating the “current winner” as it goes.
min(nums): Works identically to max, but tracks the lowest value found.
Algorithm Efficiency: These functions only need to pass through the list once, making them very fast even for lists with millions of items.

Exercise 11. Removing Duplicates from a List

Practice Problem: Write a script that takes a list containing duplicate items and returns a new list with only unique elements.

Exercise Purpose: This exercise teaches “Data De-duplication.” In real-world data science, datasets are often “messy” with repeating entries. Mastering the conversion between Lists (which allow duplicates) and Sets (which do not) is the fastest way to clean data.

Given Input: data = [1, 2, 2, 3, 4, 4, 4, 5]

Expected Output: Unique List: [1, 2, 3, 4, 5]

Hint

The easiest way to remove duplicates in Python is to convert the list into a set(), then convert it back into a list().

Solution

data = [1, 2, 2, 3, 4, 4, 4, 5]

# Set conversion removes duplicates automatically
unique_data = list(set(data))

print(f"Unique List: {unique_data}")Code language: Python (python)

Explanation to Solution:

set(data): A Set is a collection where every element must be unique. When you pass a list into a set, Python automatically discards any value it has already seen.
list(...): Since sets are unordered and do not support indexing, we convert the result back into a list so we can use it in standard list operations later.
Ordering Note: Converting to a set usually loses the original order of items. If order matters, you would need a loop or a dict.fromkeys() approach.

Exercise 12. List Comparison and Boolean Logic

Practice Problem: Write a function to return True if the first and last number of a given list is the same. If the numbers are different, return False.

Exercise Purpose: This exercise introduces “Collection Indexing” and “Boolean Flags.” Comparing data structure boundaries is common in pattern matching and data integrity checks.

Given Input:

numbers_x = [10, 20, 30, 40, 10]
numbers_y = [75, 65, 35, 75, 30]

Expected Output:

Given list: [75, 65, 35, 75, 30] | result is False
Given list: [10, 20, 30, 40, 10] | result is True

Hint

Use index 0 to access the first element and index -1 to access the very last element of the list.

Solution

def first_last_same(number_list):
    print("Given list:", number_list)
    
    first_num = number_list[0]
    last_num = number_list[-1]
    
    if first_num == last_num:
        return True
    else:
        return False

numbers_x = [10, 20, 30, 40, 10]
print("result is", first_last_same(numbers_x))

numbers_y = [75, 65, 35, 75, 30]
print("result is", first_last_same(numbers_y))Code language: Python (python)

Explanation to Solution:

Negative Indexing: Python allows [-1] to represent the last item in a list regardless of the list’s length. This is safer and cleaner than calculating len(list) - 1.
Equality Operator (==): This operator compares the values of the two objects and evaluates to a Boolean (True/False).
Logic Gate: The function provides a clear binary answer based on the structural properties of the input list.

Exercise 13. Filtering Lists with Conditional Logic

Practice Problem: Iterate through a given list of numbers and print only those numbers which are divisible by 5.

Exercise Purpose: This exercise teaches the use of the modulo operator (%) and loop filtering. In data processing, you often need to sift through large datasets to extract subsets that meet mathematical criteria.

Given Input: num_list = [10, 20, 33, 46, 55]

Expected Output:

Divisible by 5:
10, 20, 55

Hint

Use a for loop to go through each number.
Inside the loop, check if number % 5 == 0. If the remainder is zero, the number is divisible by 5.

Solution

num_list = [10, 20, 33, 46, 55]
print("Given list is", num_list)
print("Divisible by 5:")

# Iterate through each element
for num in num_list:
    # Check divisibility
    if num % 5 == 0:
        print(num)Code language: Python (python)

Explanation to Solution:

for num in num_list: This construct allows the program to visit every item in the collection without needing to track indices manually.
num % 5 == 0: The modulo operator returns the remainder. If the remainder of division by 5 is 0, the number is a multiple of 5.
Filtering Logic: Only when the if condition evaluates to True does the print function execute, effectively “filtering” the list.

Also, try to solve Python list Exercise

Exercise 14. Substring Frequency Analysis

Practice Problem: Write a program to find how many times the substring “Emma” appears in a given string.

Exercise Purpose: Text analysis and pattern matching are core pillars of programming. This exercise introduces searching for a “needle in a haystack,” a fundamental concept for building search engines or data validation tools.

Given Input:

str_x = "Emma is good developer. Emma is a writer"

Expected Output: Emma appeared 2 times

Hint

Python strings have a built-in method called .count() that can search for occurrences of a specific sequence of characters.

Solution

str_x = "Emma is good developer. Emma is a writer"
count = str_x.count("Emma")
print(f"Emma appeared {count} times")Code language: Python (python)

Explanation to Solution:

str_x.count("Emma"): This is a high-level abstraction that handles the complex logic of scanning the string and incrementing a counter internally.
Case Sensitivity: Note that string methods like count are case-sensitive; searching for “emma” (lowercase) would return 0.

Exercise 15. Nested Loops for Pattern Generation

Practice Problem: Print the following pattern where each row contains a number repeated a specific number of times based on its value.

Exercise Purpose: Pattern printing is a classic way to learn “Nested Loops.” You coordinate an outer loop for rows and an inner loop for columns or repetitions. This improves spatial logic and control over output formatting.

Given Input: Range: 1 to 5

Expected Output: (The pattern shown above)

Refer:

Hint

The outer loop should iterate from 1 to 5.
The inner loop should repeat the current number of the outer loop based on its value (e.g., when the outer number is 3, the inner loop runs 3 times).

Solution

# Outer loop for rows
for num in range(1, 6):
    # Inner loop for repetition
    for i in range(num):
        print(num, end=" ") # end=" " keeps it on the same line
    # New line after each row
    print("\n")Code language: Python (python)

Explanation to Solution:

Notice that each row contains the same number repeated, and the number of repetitions increases with the row number.

Nested Loops: The outer loop num sets the “context” for the row. The inner loop i performs the work for that specific row.
end=" ": By default, print() adds a newline. Overriding this with a space allows multiple numbers to appear side-by-side.
Formatting: The final print("\n") acts as a “carriage return,” starting the next row of the pattern on a clean line.

Exercise 16. Numerical Palindrome Check

Practice Problem: Write a program to check if a given number is a palindrome (reads the same forwards and backwards).

Exercise Purpose: This exercise introduces the idea of “Reversing Logic.” Reversing a string is simple, but reversing an integer takes some math, like using division and modulo, or changing its type. This shows how data types can work differently.

Given Input:

Case 1: number = 121
Case 2: number = 125

Expected Output:

Number 125 is not palindrome number
Number 121 is palindrome number

Refer: Python Programs to Check Palindrome Number

Hint

Use a simple “Pythonic” way str(number) to convert the number to a string and check if str_num == str_num[::-1].

Solution

def check_palindrome(number):
    print("original number", number)
    # Convert to string to reverse easily
    original_str = str(number)
    reversed_str = original_str[::-1]
    
    if original_str == reversed_str:
        print("Yes. given number is palindrome number")
    else:
        print("No. given number is not palindrome number")

check_palindrome(121)
check_palindrome(125)Code language: Python (python)

Explanation to Solution:

str(number): Type casting is used here to transform a math object into a sequence of characters.
[::-1]: This is the slice notation for reversing a sequence. It tells Python to step through the entire string backwards.
Boolean Comparison: The == operator determines if the two states (original vs. mirror) are identical.

Exercise 17. Merging Lists with Parity Filtering

Practice Problem: Create a new list from two given lists such that the new list contains odd numbers from the first list and even numbers from the second list.

Given Input:

list1 = [10, 20, 25, 30, 35]
list2 = [40, 45, 60, 75, 90]

Expected Output: [25, 35, 40, 60, 90]

Note: Try to solve the Python list exercises

Hint

Create an empty result list.
Use two separate loops to iterate through each source list and append only the numbers that satisfy the odd/even condition.

Solution

def merge_list(list1, list2):
    result_list = []
    
    # Get odd numbers from list1
    for num in list1:
        if num % 2 != 0:
            result_list.append(num)
            
    # Get even numbers from list2
    for num in list2:
        if num % 2 == 0:
            result_list.append(num)
            
    return result_list

list1 = [10, 20, 25, 30, 35]
list2 = [40, 45, 60, 75, 90]
print("result list:", merge_list(list1, list2))Code language: Python (python)

Explanation to Solution:

result_list = []: Initializing an empty list is a standard pattern for gathering data dynamically.
num % 2 != 0: This checks for “not divisible by 2,” effectively finding odd numbers.
append(): This method adds elements to the end of the new list, preserving the order in which they were found in the source lists.

Exercise 18. Integer Digit Extraction and Reversal

Practice Problem: Write a program to extract each digit from an integer in the reverse order.

Exercise Purpose: This exercise explores “Mathematical Parsing.” Instead of converting a number to a string, use the modulo operator (%) and floor division (//) to isolate digits. This is common in low-level programming and algorithm challenges where type conversion is restricted.

Given Input: number = 7536

Expected Output: 6 3 5 7

Refer: Python Programs to Reverse an Integer Number

Hint

Use number % 10 to get the last digit.
Then, use number // 10 to “chop off” that last digit and repeat the process in a loop until the number becomes zero.

Solution

number = 7536
print("Given Number:", number)

while number > 0:
    # Get the last digit
    digit = number % 10
    
    # Remove the last digit from number
    number = number // 10
    
    print(digit, end=" ")Code language: Python (python)

Explanation to Solution:

number % 10: This operation returns the remainder of the number divided by 10, which is always the rightmost digit.
number // 10: Floor division removes the decimal part, effectively shifting the number one decimal place to the right.
end=" ": This keeps the output on a single line, separated by spaces, rather than printing each digit on a new line.

Exercise 19. Multi-Tiered Income Tax Calculation

Practice Problem: Calculate income tax for a given income based on these rules:

First $10,000: 0% tax
Next $10,000: 10% tax
Remaining income: 20% tax

Exercise Purpose: This exercise introduces “Tax Brackets” logic, a classic example of complex conditional branching. It shows how to calculate values cumulatively instead of applying a single percentage to the entire amount.

Given Input: income = 45000

Expected Output: Total income tax to pay is 6000

Hint

Subtract the brackets one by one.
For an income of 45,000: the first 10k is free, the next 10k is taxed at 10% (1,000), and the remaining 25k is taxed at 20% (5,000).

Solution

income = 45000
tax_payable = 0
print("Given income:", income)

if income <= 10000:
    tax_payable = 0
elif income <= 20000:
    # Tax on first 10k is 0. Tax on the rest is 10%
    tax_payable = (income - 10000) * 10 / 100
else:
    # First 10,000 (0% tax)
    # Next 10,000 (10% tax = 1,000)
    tax_payable = 0 + (10000 * 10 / 100) 
    # Remaining income (20% tax)
    tax_payable += (income - 20000) * 20 / 100

print("Total income tax to pay is", tax_payable)Code language: Python (python)

Explanation to Solution:

Cumulative Logic: The code doesn’t just check one condition; it accounts for every bracket the income “passes through.”
elif chain: This ensures that only the relevant calculation block is executed based on the total income range.
Mathematical Precision: By breaking the calculation into steps, you avoid the common mistake of taxing the entire 45,000 at the highest rate.

Exercise 20. Nested Loops for Multiplication Tables

Practice Problem: Print a multiplication table from 1 to 10 in a formatted grid.

Exercise Purpose: To master “Matrix Generation.” This builds on the nested loop concepts from Exercise 8 and applies them to generate a structured data table. This is essential for understanding how to populate 2D arrays or generate spreadsheets.

Given Input: Range: 1 to 10

Expected Output:

1  2  3  4  5  6  7  8  9  10 		
2  4  6  8  10 12 14 16 18 20 		
... (up to 10)

Refer:

Hint

Use an outer loop for the rows (1 to 10) and an inner loop for the columns (1 to 10).
Inside the inner loop, multiply the two loop variables.

Solution

for i in range(1, 11):
    for j in range(1, 11):
        # Print product followed by a tab space
        print(i * j, end="\t")
    print("\n")Code language: Python (python)

Explanation to Solution:

range(1, 11): Remember that Python’s range is exclusive of the stop value, so we use 11 to include 10.
\t (Tab Character): This is a string escape sequence that ensures the numbers align in neat columns regardless of whether they are one or two digits long.
Row/Column Coordination: For every iteration of the outer loop (i), the inner loop (j) completes a full cycle of 1 to 10.

Exercise 21. Downward Half-Pyramid Pattern

Practice Problem: Print a downward half-pyramid pattern using stars (*).

Exercise Purpose: Learn about reverse indexing. Controlling loop boundaries in reverse is important for algorithms that process data from end to beginning.

Given Input: Rows: 5

Expected Output:

* * * * * 
* * * * 
* * * 
* * 
*

Refer:

Hint

Use a loop that starts at 5 and ends at 0, moving backwards by -1.
In each step, print the star character multiplied by the current loop value.

Solution

# Loop from 5 down to 1
for i in range(5, 0, -1):
    for j in range(0, i):
        print("*", end=" ")
    print("\n")Code language: Python (python)

Explanation to Solution:

range(5, 0, -1): The third argument -1 is the “step.” It tells the loop to decrement the value of i in each iteration.
String Multiplication (Alternative): In Python, you could also write print("* " * i), which is a more concise “Pythonic” way to repeat characters.
Inner Loop Constraint: The inner loop’s range is bound by the current value of the outer loop, causing the line length to decrease over time.

Exercise 22. Custom Exponentiation Function

Practice Problem: Write a function called exponent(base, exp) that returns an integer value of the base raised to the power of the exponent.

Exercise Purpose: Learn about “Accumulator Patterns.” Although Python has a built-in power operator (**), making your own version shows how repeated multiplication works and how functions return results to the main program.

Given Input: base = 2, exp = 5

Expected Output: 2 raises to the power of 5: 32

Hint

Initialize a result variable to 1.
Use a loop that runs exp times, and in each iteration, multiply the result by the base.

Solution

def exponent(base, exp):
    num = exp
    result = 1
    # Repeat multiplication 'exp' times
    while num > 0:
        result = result * base
        num = num - 1
    print(base, "raises to the power of", exp, "is:", result)

exponent(2, 5)
exponent(5, 4)Code language: Python (python)

Explanation to Solution:

Base Case: result starts at 1 because 1 is the identity element for multiplication (anything multiplied by 1 remains itself).
The while Loop: This controls the number of multiplications. Each cycle represents one “power.”
Function Reusability: By passing base and exp as parameters, the same block of code can calculate 2⁵, 5⁴, or any other combination.

Exercise 23. Check Palindrome Number

Practice Problem: Write a program to check if a given number is a palindrome. A palindrome number remains the same when its digits are reversed (e.g., 121, 545).

Exercise Purpose: This exercise teaches “Algorithmic Reversal.” While strings are easy to reverse in Python, reversing a number mathematically using the modulo (%) and floor division (//) operators deepens understanding of how integers are stored in memory and how to manipulate digits individually.

Given Input: number = 121

Expected Output:

Original number 121
Yes. given number is palindrome number

Refer:

Hint

You can reverse the number by repeatedly extracting the last digit (num % 10) and building a new number.
Alternatively, for a “Pythonic” shortcut, convert the integer to a string and use slicing [::-1] to compare it to its original form.

Solution

def check_palindrome(number):
    # Convert to string to easily reverse
    str_num = str(number)
    reverse_str = str_num[::-1]
    
    if str_num == reverse_str:
        print(f"Original number {number}")
        print("Yes. given number is palindrome number")
    else:
        print(f"Original number {number}")
        print("No. given number is not palindrome number")

check_palindrome(121)Code language: Python (python)

Explanation to Solution:

str(number): Converts the integer into a sequence of characters so we can treat it like a list.
[::-1]: This is the slice notation for “start at the end, end at the beginning, and move backwards by 1.” It effectively mirrors the string.
if str_num == reverse_str: A boolean comparison that checks for exact symmetry.

Exercise 24. Generate Fibonacci Series

Practice Problem: Write a program to print the first 15 terms of the Fibonacci series. The sequence starts with 0 and 1, and each subsequent number is the sum of the two preceding ones.

Exercise Purpose: The Fibonacci sequence is a classic way to learn about state management in loops. You keep track of two changing variables at once to find the next number, which helps you see how data moves through each step.

Given Input: Terms = 15

Expected Output: 0 1 1 2 3 5 8 13 21 34 55 89 144 233 377

Refer: Generate Fibonacci Series in Python

Hint

Initialize two variables, num1 = 0 and num2 = 1.
In each iteration of a loop, the next number is num1 + num2.
Then, update num1 to be num2, and num2 to be the new sum.

Solution

num1, num2 = 0, 1
print("Fibonacci series:")

for i in range(15):
    print(num1, end="  ")
    # Calculate next term
    res = num1 + num2
    # Update values for next iteration
    num1 = num2
    num2 = resCode language: Python (python)

Explanation to Solution:

num1, num2 = 0, 1: Python’s tuple unpacking allows for clean initialization of multiple variables.
res = num1 + num2: This creates the new term based on the current state of the two pointers.
num1 = num2 and num2 = res: This “shifts” the window forward by one position, preparing the variables for the next cycle.

Exercise 25. Check Leap Year

Practice Problem: Write a program that takes a year as input and determines if it is a leap year.

A leap year is a year in the Gregorian calendar that contains an extra day, making it 366 days long instead of the usual 365. This extra day, February 29th, is added to keep the calendar synchronized with the Earth’s revolution around the Sun.

Rules for leap years: a year is a leap year if it’s divisible by 4, unless it’s also divisible by 100 but not by 400.

Exercise Purpose: This exercise is vital for mastering “Complex Conditional Logic.” A leap year isn’t just “every 4 years”; there are specific exceptions for century years. This forces the programmer to use nested if statements or compound logical operators (and/or).

Given Input: year = 2024

Expected Output: 2024 is a leap year

Hint

A year is a leap year if it is divisible by 4. However, if it is divisible by 100, it is NOT a leap year UNLESS it is also divisible by 400.

Solution

def is_leap(year):
    # Standard Leap Year Logic
    if (year % 4 == 0 and year % 100 != 0) or (year % 400 == 0):
        print(f"{year} is a leap year")
    else:
        print(f"{year} is not a leap year")

is_leap(2024)
is_leap(2100)Code language: Python (python)

Explanation to Solution:

year % 4 == 0: Checks the basic 4-year cycle.
year % 100 != 0: Excludes years like 1900 or 2100, which are divisible by 4 but are not leap years.
or (year % 400 == 0): The “exception to the exception,” ensuring years like 2000 are correctly identified as leap years.

Exercise 26. Merging Two Dictionaries

Practice Problem: Write a program that takes two separate dictionaries and merges them into one single dictionary.

Exercise Purpose: This introduces “Key-Value Consolidation.” Merging dictionaries is a common task when combining configuration files or user profiles. It also teaches you about “Key Overwriting”—what happens when both dictionaries share the same key.

Given Input:

dict1 = {"name": "Alice", "age": 25}
dict2 = {"city": "New York", "job": "Engineer"}Code language: Python (python)

Expected Output:

{'name': 'Alice', 'age': 25, 'city': 'New York', 'job': 'Engineer'}

Hint

In modern Python (3.9+), you can use the union operator | to merge two dictionaries effortlessly.
For older versions, the .update() method is used.

Solution

dict1 = {"name": "Alice", "age": 25}
dict2 = {"city": "New York", "job": "Engineer"}

# Modern Python merge (Union Operator)
merged_dict = dict1 | dict2

print(merged_dict)Code language: Python (python)

Explanation to Solution:

The | Operator: This creates a new dictionary containing the keys and values of both inputs.
Conflict Resolution: If both dictionaries had an “age” key, the value from the second dictionary (dict2) would prevail in the final merge.
Immutability: This method preserves the original dict1 and dict2 while creating a third, combined object.

Exercise 27. Finding Common Elements (Intersections)

Practice Problem: Take two lists and find the elements that appear in both. Use Sets to perform the operation.

Exercise Purpose: This exercise explores “Mathematical Set Operations.” Finding intersections is vital for recommendation engines (e.g., finding “mutual friends” or “shared interests”). It demonstrates why using the right data structure (Set) is more efficient than nested loops.

Given Input:

list_a = [1, 2, 3, 4, 5]
list_b = [4, 5, 6, 7, 8]Code language: Python (python)

Expected Output: Common Elements: {4, 5}

Hint

Convert both lists to sets.
Use the & operator (Intersection) to find the elements that exist in both sets.

Solution

list_a = [1, 2, 3, 4, 5]
list_b = [4, 5, 6, 7, 8]

set_a = set(list_a)
set_b = set(list_b)

# Find common elements using intersection
common = set_a & set_b

print(f"Common Elements: {common}")Code language: Python (python)

Explanation to Solution:

set_a & set_b: The ampersand represents the “Intersection” operation. It returns a set containing only items that are present in both set_a AND set_b.
Performance: Doing this with sets is significantly faster than using nested for loops, as set lookups are nearly instantaneous (O(1) average case).
Result Type: The output is a set {4, 5}, which effectively highlights that we are looking for a unique collection of shared items.

Exercise 28. Odd/Even List Splitter

Practice Problem: Start with a list of 10 numbers. Iterate through them and sort them into two separate lists: one for even numbers and one for odd numbers.

Given Input: numbers = [12, 7, 34, 21, 5, 10, 8, 3, 19, 2]

Expected Output:

Even numbers: [12, 34, 10, 8, 2]
Odd numbers: [7, 21, 5, 3, 19]

Hint

Create two empty lists at the start.
Use a for loop combined with an if-else statement.
Remember, a number is even if num % 2 == 0.

Solution

numbers = [12, 7, 34, 21, 5, 10, 8, 3, 19, 2]
evens = []
odds = []

for num in numbers:
    if num % 2 == 0:
        evens.append(num)
    else:
        odds.append(num)

print(f"Even numbers: {evens}")
print(f"Odd numbers: {odds}")Code language: Python (python)

Explanation to Solution:

% 2 == 0: The modulo operator checks for a remainder. If a number divided by 2 has no remainder, it is mathematically even.
.append(): This method dynamically grows our target lists as the loop finds matching candidates.
Logical Branching: The if-else structure ensures that every number is placed in exactly one category, maintaining the integrity of the original dataset.

Exercise 29. Word Length Analysis

Practice Problem: Create a list of 5 words. Write a loop that iterates through the list and prints each word alongside its character count.

Exercise Purpose: This exercise introduces “Metadata Extraction.” Often, you aren’t just interested in the data itself, but in its properties. In web development, this logic is used to validate if a user’s password or username meets specific length requirements.

Given Input: words = ["Apple", "Banana", "Cherry", "Date", "Elderberry"]

Expected Output:

Apple - 5 Banana - 6 Cherry - 6 Date - 4 Elderberry - 10

Hint

Use a for loop to access each word and the built-in len() function to calculate the number of letters.

Solution

words = ["Apple", "Banana", "Cherry", "Date", "Elderberry"]

for word in words:
    length = len(word)
    print(f"{word} - {length}")Code language: Python (python)

Explanation to Solution:

len(): This is a universal Python function that returns the number of items in a sequence (in this case, characters in a string).
String Interpolation (f-strings): We use f"{word} - {length}" to create a clean, readable output format that combines variables and static text.
Sequential Processing: The loop ensures that the operation is performed on every item in the list automatically, regardless of how many words are added later.

Exercise 30. Word Frequency Counter (The Histogram)

Practice Problem: Write a program that counts how many times each word appears in a given paragraph and stores these counts in a dictionary.

Exercise Purpose: This is a classic “Natural Language Processing” (NLP) task. It teaches you how to map data to occurrences, which is the logic used by search engines to index web pages or by social media platforms to identify trending hashtags.

Given Input: text = "apple banana apple cherry banana apple"

Expected Output: {'apple': 3, 'banana': 2, 'cherry': 1}

Hint

Split the text into a list of words.
Iterate through the list; if a word is already in the dictionary, increase its value by 1.
If it isn’t, add it with a starting value of

Solution

text = "apple banana apple cherry banana apple"
words = text.split()
frequency = {}

for word in words:
    if word in frequency:
        frequency[word] += 1
    else:
        frequency[word] = 1

print(frequency)Code language: Python (python)

Explanation to Solution:

.split(): This breaks the string into a list of individual strings, allowing us to process them one by one.
Dictionary Membership: if word in frequency checks if the word has already been “seen” by the script.
Incrementing Values: By using the word as a key and the count as a value, we create a high-speed lookup table of the paragraph’s content.

Exercise 31. Print Alternate Prime Numbers

Practice Problem: Write a program to find all prime numbers up to 20, but only print every second (alternate) prime number found.

Exercise Purpose: This exercise combines “Nested Loops” (to check for primality) with “Step Logic.” It requires the programmer to first identify a subset of data and then apply a secondary filter, a common task in data reporting.

Given Input: Limit = 20

Expected Output: 2, 5, 11, 17

Refer:

Hint

First, create a list of all primes between 2 and 20.
Once the list is complete, use list slicing with a step of 2 (primes[::2]) to print the alternate values.

Solution

primes = []

for num in range(2, 21):
    # Check if number is prime
    for i in range(2, int(num**0.5) + 1):
        if num % i == 0:
            break
    else:
        primes.append(num)

# Print alternate primes
alternate_primes = primes[::2]
print(alternate_primes)Code language: Python (python)

Explanation to Solution:

for...else loop: A unique Python feature where the else block executes only if the for loop completes without hitting a break. This is perfect for prime checking.
num**0.5: Efficiency trick—you only need to check factors up to the square root of a number.
primes[::2]: The slicing engine skips every other element in the list, fulfilling the “alternate” requirement

Exercise 32. Dictionary of Squares (Mapping Logic)

Practice Problem: Create a dictionary where the keys are numbers from 1 to 10 and the values are the squares of those numbers (e.g., 2: 4, 3: 9).

Exercise Purpose: This exercise explores “Data Mapping.” It demonstrates how dictionaries can be used to store pre-calculated mathematical relationships, essentially acting as a “lookup table” that can replace expensive repetitive calculations.

Given Input: Range: 1 to 10

Expected Output:

{1: 1, 2: 4, 3: 9, 4: 16, 5: 25, 6: 36, 7: 49, 8: 64, 9: 81, 10: 100}

Hint

Use range(1, 11) and a loop to populate the dictionary.
You can assign values using the syntax dict[key] = value.

Solution

squares = {}

for i in range(1, 11):
    squares[i] = i * i

print(squares)Code language: Python (python)

Explanation to Solution:

range(1, 11): This provides the “keys” for our dictionary.
i * i: This calculates the square of the current number.
Key-Value Assignment: The line squares[i] = i * i creates a permanent link between the input number and its squared result within the dictionary structure.

Exercise 33. Character Replacer (Data Sanitization)

Practice Problem: Ask the user for a sentence. Replace every empty space in that sentence with an underscore (_) and print the final result.

Exercise Purpose: This exercise focuses on “String Sanitization.” In web development and file management, spaces are often problematic (especially in URLs or file paths). Learning to replace characters is a critical skill for preparing data for storage or transmission.

Given Input: "I love coding in Python"

Expected Output: I_love_coding_in_Python

Hint

Python strings have a built-in method called .replace(old, new) that scans the entire string and swaps characters for you.

Solution

user_sentence = input("Enter a sentence: ")

# Replace space with underscore
sanitized_sentence = user_sentence.replace(" ", "_")

print(sanitized_sentence)Code language: Python (python)

Explanation to Solution:

.replace(" ", "_"): This method is “Global,” meaning it doesn’t just find the first space; it finds every space in the string and replaces it.
Immutability: Remember that user_sentence itself isn’t changed; replace() creates a new string which we store in sanitized_sentence.
Input Handling: By using input(), the program can handle any sentence the user provides, making it a flexible utility script.

Exercise 34. Print Reverse Number Pattern

Practice Problem: Print a downward number pattern where each row starts with a decreasing value.

Exercise Purpose: In this exercise, you will learn about range control and practice using negative steps in loops to move backwards. This skill is important for algorithms that process data from the end of a file to the beginning.

Given Input: Rows = 5

Expected Output:

Refer:

Hint

You need two loops. The outer loop should start at 5 and go down to 1.
The inner loop should start at the current value of the outer loop and go down to 1.

Solution

rows = 5
# Outer loop for number of rows
for i in range(rows, 0, -1):
    # Inner loop for printing numbers in each row
    for j in range(i, 0, -1):
        print(j, end=' ')
    print("") # New lineCode language: Python (python)

Explanation to Solution:

range(rows, 0, -1): The -1 argument is the step. It tells Python to count backwards.
print(j, end=' '): By overriding the default newline with a space, we keep the numbers on the same horizontal line.
print(""): This simple empty print statement acts as a “carriage return” to start the next row of the pattern.

Exercise 35. Digit Detection in Strings

Practice Problem: Write a program to check if a user-entered string contains any numeric digits. Use a for loop to examine each character.

Exercise Purpose: This exercise will help you learn about string traversal and character analysis. In software development, these skills are important for tasks like checking if a username has forbidden characters or if a password is complex enough.

Given Input: input_string = "Python3"

Expected Output: The string 'Python3' contains digits: True

Hint

Iterate through the string using a for loop.
For each character, use the built-in .isdigit() method.
If you find even one digit, you can set a flag to True and break the loop.

Solution

user_input = "Python3"
contains_digit = False

# Iterate through each character
for char in user_input:
    if char.isdigit():
        contains_digit = True
        break  # Exit early since we found one

print(f"The string '{user_input}' contains digits: {contains_digit}")Code language: Python (python)

Explanation to Solution:

Flag Pattern: We initialize contains_digit as False. This “flag” only flips if a specific condition is met.
.isdigit(): This character method returns True if the character is a numeric value (0-9) and False otherwise.
break: This keyword is an efficiency tool. Once a single digit is found, there is no need to check the remaining characters, saving processing time.

Exercise 36. Capitalize First Letter (Title Case)

Practice Problem: Write a program to capitalize the first letter of each word in a given string without using the built-in .title() method.

Exercise Purpose: In this exercise, you will learn about “Tokenization” and “String Re-assembly.” You will split a sentence into words, change them, and then put the sentence back together. This helps you practice working with complex data structures.

Given Input: text = "hello world from python"

Expected Output: Hello World From Python

Hint

Use the .split() method to turn the string into a list of words.
Loop through the list, use .capitalize() on each word, and then use " ".join() to merge them back into a single sentence.

Solution

text = "hello world from python"

# Split the string into a list of words
words = text.split()
capitalized_words = []

for word in words:
    # Capitalize each word and add to the new list
    capitalized_words.append(word.capitalize())

# Join the list back into a single string
result = " ".join(capitalized_words)
print(result)Code language: Python (python)

Explanation to Solution:

.split(): By default, this breaks a string into a list wherever there is a space. "hello world" becomes ["hello", "world"].
.capitalize(): This method turns the first character of a string to uppercase and the rest to lowercase.
" ".join(): This is the inverse of split. It takes a list and glues the items together using the string it is called on (in this case, a space) as the adhesive.

Exercise 37. Simple Countdown Timer

Practice Problem: Create a countdown timer that starts from a given number and counts down to zero using a while loop.

Exercise Purpose: In this exercise, you will learn about loop termination logic and time delay management. Knowing how to control the flow of your code in real time is important for making animations, game loops, or automated scripts.

Given Input: start_count = 5

Expected Output: 5 4 3 2 1 Blast off!

Hint

Use a while loop that continues as long as the counter is greater than 0.
Inside the loop, print the current value and then subtract 1 from the counter.

Solution

import time

count = 5

while count > 0:
    print(count)
    # Pause the program for 1 second
    time.sleep(1)
    # Decrement the counter
    count -= 1

print("Blast off!")Code language: Python (python)

Explanation to Solution:

while count > 0: This condition ensures the loop runs exactly as many times as specified. Once count hits 0, the condition becomes False and the loop stops.
time.sleep(1): This function from the time module pauses execution for one second, making the output feel like a real clock.
count -= 1: This is shorthand for count = count - 1. Without this line, the loop would be “infinite” because count would always stay at 5.

Exercise 38. File Creation and Basic I/O

Practice Problem: Write a program that creates a new text file named notes.txt, writes three separate lines of text to it, and then reads that file back to display the contents in the console.

Exercise Purpose: This exercise introduces “Persistent Storage.” Unlike variables that disappear when the program stops, files allow you to save data to the hard drive. Learning the open(), write(), and read() workflow is essential for building logging systems and saving user settings.

Given Input: Lines to write:

“Hello, this is my first note.”
“Python file handling is simple.”
“End of file.”

Expected Output:

notes.txt

Hello, this is my first note.
Python file handling is simple.
End of file.

Hint

Use the with open("notes.txt", "w") statement to ensure the file is properly closed after writing.
Use the "w" mode for writing and the "r" mode for reading.

Solution

# Part 1: Writing to the file
with open("notes.txt", "w") as file:
    file.write("Hello, this is my first note.\n")
    file.write("Python file handling is simple.\n")
    file.write("End of file.\n")

# Part 2: Reading from the file
print("Reading file contents:")
with open("notes.txt", "r") as file:
    content = file.read()
    print(content)Code language: Python (python)

Explanation to Solution:

with open(...): Known as a “Context Manager,” this ensures that Python automatically closes the file even if an error occurs, preventing memory leaks or file corruption.
\n: This is the “newline” character. Without it, all three sentences would be squashed together on a single line in the text file.
"w" vs "r": The mode "w" (Write) will overwrite the file if it already exists, while "r" (Read) opens it for viewing only.

Exercise 39. External File Word Counter

Practice Problem: Write a script that opens an existing .txt file and counts the total number of words it contains.

Exercise Purpose: This exercise teaches “Data Parsing.” In professional environments, you rarely work with data you typed into the code yourself; you almost always pull data from external sources. This script simulates basic text-mining techniques used to analyze documents or logs.

Given Input:

An external file sample.txt containing: “Coding is the language of the future.”

Expected Output: The file contains 7 words.

Hint

Read the entire content of the file into a string, then use the .split() method to break that string into a list of words.
The length of that list is your word count.

Solution

# Assuming 'sample.txt' exists with some text
try:
    with open("sample.txt", "r") as file:
        data = file.read()
        words = data.split()
        word_count = len(words)
        print(f"The file contains {word_count} words.")
except FileNotFoundError:
    print("Error: The file 'sample.txt' was not found.")Code language: Python (python)

Explanation to Solution:

data.split(): By default, split() breaks a string at every space or newline. This effectively creates a list where every element is a single word.
len(words): Since words is now a list, the length of the list directly corresponds to the count of words in the document.
try-except: This is a bonus safety feature. If the file is missing, the program won’t “crash” with a scary error; instead, it will print a polite message explaining what happened.

Exercise 40. Introduction to Classes (OOP)

Practice Problem: Create a Car class with attributes for make, model, and year. Include a method called start_engine() that prints a formatted string describing the car starting up.

Exercise Purpose: This exercise introduces Object-Oriented Programming (OOP). Instead of just writing functions, you are creating a “Blueprint” (the Class) to generate “Objects” (the specific cars). This is how modern software is built, allowing you to organize code into logical, reusable components.

Given Input: Make: “Toyota”, Model: “Camry”, Year: 2022

Expected Output: The 2022 Toyota Camry's engine is now running!

Hint

Use the __init__ method to initialize your attributes and the self keyword to refer to the specific instance of the car being used.

Solution

class Car:
    def __init__(self, make, model, year):
        # Setting up attributes
        self.make = make
        self.model = model
        self.year = year

    def start_engine(self):
        # A method that uses the object's attributes
        print(f"The {self.year} {self.make} {self.model}'s engine is now running!")

# Creating an object (an instance of the class)
my_car = Car("Toyota", "Camry", 2022)

# Calling the method
my_car.start_engine()Code language: Python (python)

Explanation to Solution:

__init__: This is the “Constructor.” It runs automatically the moment you create a new car, setting its identity (Toyota, Camry, etc.).
self: This is a placeholder for the specific object. It tells Python, “use this car’s year” rather than just a generic variable named year.
Encapsulation: By grouping the data (attributes) and the actions (methods) together, your code becomes modular and much easier to manage as your program grows.

Comments

Abhinay Soni says

March 1, 2026 at 1:04 pm

Thank you so much Vishal Hule, for creating this amazing content, it help a lot.

Reply
frank carlsen says

February 22, 2026 at 10:51 pm

Exercise 39. It seems that the file sample.txt is no longer avaliable or was it meant to be created?

Traceback (most recent call last):
File “/home/repl961/main.py”, line 3, in
with open(“sample.txt”, “r”) as file:
^^^^^^^^^^^^^^^^^^^^^^^
FileNotFoundError: [Errno 2] No such file or directory: ‘sample.txt’

Reply
- Abhinay Soni says
  
  March 1, 2026 at 12:59 pm
  
  Yes, you need to create the sample.txt before you try to read from it
  
  Reply
Octivius says

January 29, 2026 at 11:38 am

for number 16 i suggest this alternative
def palindrome(num):
numeral = num
num_s = str(num)
num_r = int(num_s[::-1])
if numeral == num_r:
print(f”The number {num} is a palindrome”)
else:
print(f”The number {num} is not a palindrome”)
palindrome(525)
palindrome(4554)
palindrome(4543)

Reply
Prosper Ekhator says

January 28, 2026 at 1:14 am

for exercise 2.. i think this is the best way to do it,,,,

def first_ten_num(num):
print(f’Printing current and previous number sum in a range({num})’)
previous_num = 0
for i in range(num):
current_num = i
sum_num = current_num + previous_num
print(f’Current Number {current_num} Previous Number {previous_num} Sum: {sum_num}’)
previous_num = current_num

Reply
Suyash Mishra says

December 19, 2025 at 12:51 pm

this is my code for 3rd question:

n = input()
for i in n[::2]:
print(i)

I am new to code ,pls tell

Reply
Ufok says

November 3, 2025 at 3:24 am

Exercise 18. Don’t ask me how much time it took me to consider casting to bool xD

def is_leap(y):
return bool(not y % 4 and y % 100 or not y % 400)

print(‘*** True’)
print(is_leap(2400))
print(is_leap(2404))
print(‘*** False’)
print(is_leap(2100))
print(is_leap(2401))

Reply
Mehdi Parsa says

October 21, 2025 at 6:32 pm

Exercise 20 >>
a=int(input(‘Enter Range :’))
for i in range(a, 0, -1):
print((str(i)+’ ‘)*i+’\n’)

Reply
Minhaj Shaikh says

July 17, 2025 at 9:08 pm

Hey Vishal there is an alternate method of ex.5

list1 = [10, 20, 25, 30, 35]
list2 = [40, 45, 60, 75, 90]
l3=[]
for i in range(len(list1)):
if list1[i]%2!=0:
l3.append(list1[i])
if list2[i]%2==0:
l3.append(list2[i])
print(l3)

list1 = [10, 20, 25, 30, 35]
list2 = [40, 45, 60, 75, 90]
l3=[]
#print(list(l3))
for a,b in zip(list1,list2):
if a%2!=0:
l3.append(a)
if b%2==0:
l3.append(b)
print(l3)

Reply
Sai Ganesh J says

June 26, 2025 at 11:05 pm

this seems to be a much simple and generalised code, am i wrong vishal?
a = int(input(“Enter the digit to be reversed : “))
b = str(a)
for i in range(len(b)) :
print(b[len(b)-i-1],” “,end=”)

Reply
Sai Ganesh J says

June 26, 2025 at 11:02 pm

for the 11th question this seems to be a simple method and generalised one too, am i wrong ?
a = int(input(“Enter the digit to be reversed : “))
b = str(a)
for i in range(len(b)) :
print(b[len(b)-i-1],” “,end=”)

Reply
Gijs Wiersema says

June 17, 2025 at 9:04 pm

For excersise 8, I actually did it with a single for loop:

def numPyramid(count):
for i in range(count+1):
print((str(i)+’ ‘)*i + ‘\n’)
numPyramid(5)

I love doing the excersises by the way. It’s a great way to refresh my python knowledge.

Reply
Yas says

March 31, 2025 at 12:42 am

Exercise 9

number_string = input(“Enter the number : “)
number_string_reversed = number_string [::-1]
int(number_string_reversed)

if number_string_reversed==number_string :
print(“Yes. given number is palindrome number”)
else :
print(“No. given number is not palindrome number”)

Reply
Yas says

March 31, 2025 at 12:07 am

Exercise 3
name = input(“Enter a string : “)
x = len(name)
for i in name :
print(i)

Reply
- kp singh says
  
  April 9, 2025 at 4:17 pm
  
  you take a string and after then calculate the length when you use loop then one by one string character print on scrren
  
  Reply
Srinivas says

December 25, 2024 at 12:31 pm

income=int(input(“Income: “))
taxable_income = (income – 10000)

if income<=10000:
print("No Tax")
elif (0<taxable_income10000:
slab1 = 10000
slab2 = taxable_income – 10000
total_tax= (slab1*10/100)+(slab2*20/100)
print(“Your tax amount is: “,total_tax)

Reply
- Virus says
  
  February 13, 2025 at 2:50 am
  
  income = int(input(“Income: “))
  taxable_income = income – 10000
  
  if income <= 10000:
  print("No Tax")
  elif 0 < taxable_income <= 10000:
  slab1 = 10000
  slab2 = taxable_income
  total_tax = slab1 * 10 / 100 + slab2 * 20 / 100
  print("Your tax amount is: ", total_tax)
  else:
  slab1 = 10000
  slab2 = 10000
  slab3 = taxable_income – 20000
  total_tax = (slab1 * 10 / 100) + (slab2 * 20 / 100) + (slab3 * 30 / 100)
  print("Your tax amount is: ", total_tax)
  
  Reply
Malcolm says

November 14, 2024 at 6:29 pm

These exercises are so helpful! This is the way I solved #15:

base = int(input(‘Raise ‘))
power = int(input(‘to the power of ‘))
print(‘{} raised to the power of {} is {}’.format(base, power, base**power))

Reply
- Sai says
  
  June 28, 2025 at 10:03 pm
  
  it clearly says create a function, you have created no such functions instead used something simple(not the point of solving the question, its just common sense at this point), stick to the question.
  
  Reply
sara says

October 18, 2024 at 11:31 pm

base=2
exponent=3
result1=1
for i in range(0,exponent):
result1=base*result1

print(result1)

Reply
- Carlos says
  
  October 29, 2024 at 12:29 am
  
  print (“\nExercise 15: power of the exponent”)
  
  base=2
  exponent=3
  
  print(base, “raises to the power of “, exponent, “:”, base**exponent, end=” “)
  
  print(” ( “, end=” “)
  print(base, end=” “)
  for i in range(exponent-1):
  print(” * “, base, end=” “)
  print(” = “, base**exponent, end=” “)
  print(” ) “, end=” “)
  
  OUTPUT
  
  Exercise 15: power of the exponent
  2 raises to the power of 3 : 8 ( 2 * 2 * 2 = 8 )
  
  Reply
- Sai says
  
  June 28, 2025 at 10:03 pm
  
  it clearly says create a function, you have created no such functions instead used something simple(not the point of solving the question, its just common sense at this point), stick to the question.
  
  Reply
SilentDeath_ says

September 22, 2024 at 5:05 pm

def is_palindrome(num):
str_num = str(num)
print(f”original number {str_num}”)
reversed_num = str_num[::-1]
if str_num == reversed_num:
print(“Yes. given number is palindrome number”)
else:
print(“No. given number is not palindrome”)

Reply
- Amine says
  
  September 26, 2024 at 4:29 am
  
  we can directly do that:
  def is_palindrome(n):
  if str(n) == str(n)[::-1]:
  return True
  else:
  return False
  
  print(is_palindrome(121))
  
  Reply
  - amine says
    
    November 17, 2024 at 10:24 pm
    
    We should work this exercice without changing the type to string
    
    Reply
- hacker says
  
  August 12, 2025 at 1:11 pm
  
  That’s too long use this
  
  list1 = [1, 2, 1]
  
  copy_list1 = list1.copy()
  
  copy_list1.reverse()
  
  if(copy_list1 == list1):
  print(“Palindrome”)
  else:
  print(“Not Palindrome”)
  
  Reply
SilentDeath_ says

September 22, 2024 at 4:20 pm

#Solution on question number ( 8 ):

for x in range(n):
print(f”{x}” * x) #–> The (f”{x}”) will make it a string and multiply it by the number x in the range.

#It’s move easier like this. What you all think?

Reply
- Henry says
  
  September 23, 2024 at 9:40 am
  
  for i in range(1, 6):
  print(str(i) * i)
  
  It’s all in 2 lines
  
  Reply
yaser says

August 19, 2024 at 2:42 pm

solution to the exercise number 6:

def divisable_numbers():
numbers = [10,20,34,32,40,25,245]
print(“Given list is”, numbers)
print(“Divisable numbers to 5”)
for numbers in numbers:
if numbers % 5 == 0:
print(“.”,numbers)

divisable_numbers()

Reply
yaser says

August 18, 2024 at 12:22 pm

question number 3:

word = input(“Enter your string: “)
print(“The original word is”, word)

result = word[0::2]
for _ in result:
print(_)

Reply
Hema Suresh says

August 6, 2024 at 10:57 am

Ex: 15

base = int(input(“ENTER BASE”))
exponent = int(input(“ENTER EXPONENT”))
res=1
for i in range(0,exponent,1):
res=base*res
print(res)

Reply
- Python says
  
  September 28, 2024 at 2:30 am
  
  def exponent(b,e):
  return print(pow(b,e))
  exponent(2,5)
  
  Reply
amir goodarzi says

July 30, 2024 at 7:17 pm

the answer of the second exercise is wrong 🙁
I think this answer is better than yours!
#this is a program that solve the second exercise

print(“printing current and pervious number sum in a range(10)”)

pervious = 0

for i in range(10):
pervious = i – 1
if i == 0 :
print(“current number %i pervious number %i sum: %i” %(0, 0, 0))
continue
print(“current number %i pervious number %i sum: %i” %(i, i-1, i + pervious))

Reply
amir goodarzi says

July 30, 2024 at 7:14 pm

in third exercise , better use f string instead **print(“index[“, i, “]”, word[i])**

Reply
- yaser says
  
  August 18, 2024 at 12:25 pm
  
  what about this method?
  
  word = input(“Enter your string: “)
  print(“The original word is”, word)
  
  result = word[0::2]
  for _ in result:
  print(_)
  
  Reply
Asher Sajid says

July 13, 2024 at 9:16 pm

For exercise 15 we can also do:

def exponent(base, exp):
x = pow(base, exp)
print(x)

exponent(2, 5)

Reply
Dariusz says

June 6, 2024 at 5:11 pm

task 8

for x in range(5+1):
print(f”{x} “*x)

works for my, and looks cleaner

Reply
jose placencio says

May 28, 2024 at 2:18 am

my answer for the 15 one it is this:

def exponential(base,exponen):
print((base)**(exponen))

the parenthesis if is for acepting both negative and positive numbers

Reply
Dhamo says

April 28, 2024 at 1:37 pm

Exercise 15

def exponent(base,exp):
if exp >= 1:
exponent = base ** exp
return exponent
else:
return f’You have entered a invalid value:{exp}, please enter a valid value’

Reply
Mayur says

April 27, 2024 at 2:45 am

Q.no : 15 in simple way .

def exponent(base,exp):
print(‘exponent of given numbers:’,base**exp)

exponent(2,5)

Reply
- mvelo says
  
  May 23, 2024 at 6:27 pm
  
  this is a convenient way to accommodate any number from the user according to the instructions given.
  
  ”’exponent must be positive
  the base must be an integer”
  
  def integer_check():
  while True:
  base = input(“enter a base: “)
  try:
  value = int(base)
  return value
  
  except ValueError:
  pass
  
  def exponent_check():
  while True:
  exponent = input(“enter the exponent: “)
  if exponent.isdigit():
  if int(exponent) > 1:
  return int(exponent)
  else:
  pass
  else:
  pass
  
  def expression(value, exponent):
  return value ** exponent
  
  def output():
  base = integer_check()
  exponent = exponent_check()
  result = expression(base, exponent)
  print(f’Base: {base}’)
  print(f’Exponent: {exponent}’)
  print(f”{base} raised to the power of {exponent} is: {result}”)
  print(f”)
  
  # Run the output function
  
  Reply
Kriyansh Sinha says

April 16, 2024 at 4:17 am

Ex: 14
for i in range(5,0,-1):
print(‘*’*i,sep=’ ‘, end=”\n”)

better code

Reply
Kriyansh Sinha says

April 16, 2024 at 4:09 am

Ex: 14

for i in range(5,0,-1):
print(‘*’*i,sep=’ ‘, end=”\n”)

Reply
Adesayo says

April 7, 2024 at 10:14 am

EXERCISE 13 SOLUTION

for i in range(1, 11):
for j in range(1, 11):
print((i * j), end=” “)
print()

Reply
Adesayo says

April 7, 2024 at 9:56 am

EXECIRSE 12 SOLUTION

income = int(input(“how much do you earn? “))
tax_rate = 0

if income <= 10000:
tax_rate = 0
elif income <= 20000:
tax_rate = (income-10000) * 0.1
else: # has to be over 20000
tax_rate = (10000 * 0.1) + ((income-20000) * 0.2)

print("THE TAX RATE 0N ${} IS ${}".format(income, tax_rate))

Reply
Sebas says

April 3, 2024 at 1:27 am

# Exercise 13: Print multiplication table from 1 to 10
# the pythonic way
[[print(x*y, end=” “) for x in range(1,11)] and print() for y in range(1,11)]

Reply
Narendran says

March 18, 2024 at 12:34 pm

print(“Printing current and previous number and their sum in a range(10)”)
for i in range(10):
cur = i
if i == 0:
pre = i+0
sum = i+0
else:
pre = i-1
sum = i+pre
print(f”Current Number {cur} Previous Number {pre} Sum: {sum}”)

Reply
Fziez says

February 24, 2024 at 2:31 pm

why in exercise 3 :

for i in range(0, size – 1, 2):
print(“index[“, i, “]”, word[i])

for loop end at size-1? when I append ‘h’ in pynative it never print

Reply
- Yaron says
  
  March 11, 2024 at 3:30 am
  
  Thanks man !
  Nice exercises
  
  Reply
- Dexter Varquez says
  
  February 19, 2025 at 2:23 pm
  
  user_str = input(“Enter a string: “)
  
  print(“\nOriginal String is”, user_str)
  print(“Printing only even index chars\n”)
  
  for i in range(0, len(user_str), 2):
  print(user_str[i])
  
  Reply
Vignesh says

February 20, 2024 at 7:55 pm

Excercise 15:
def exponent(base, exp):
return base ** exp

print(exponent(5,4))

Reply
- Mukul Chauhan says
  
  March 11, 2024 at 6:18 pm
  
  you dont need to print it.
  just pass
  
  exponent(5,4)
  
  Reply
Steven Shen says

February 18, 2024 at 12:21 pm

Here is my way to do Exercise 15 for matching Expected Output and as a reference to you:
————————————————————————————————–
base,exponent=map(int,input(‘Please input your base & exponent with a “,”: ‘).split(‘,’))
print(f'{base} raises to the power of {exponent} is: ‘,base**exponent)
result_list=list(str(base)*exponent)
for i in range(len(result_list)-1):
result_list[i] += ‘ *’
result_list[-1] += ‘ = ‘+str(base**exponent)
result_str=”.join(result_list)
print(‘i.e. (%s)’ %result_str)

Reply
Adam says

February 15, 2024 at 3:08 pm

My solutions were almost always different to authors’ suggestions.
1.
x = int(input(“What’s number 1 “))
y = int(input(“What’s number 2 “))
z = x*y
if z <= 1000:
print(z)
else:
print(x+y)
2.
print("Printing current and previous number sum in a range(10)")
print("Current Number 0 Previous Number 0 Sum: 0")
x = 1

while x < 10:
print("Current Number", x, "Previous Number ", x-1, " Sum: ", x+x-1)
x = x+1
3.
x = input("Orginal String is ")
print("Printing only even index chars")
for i in range(0, len(x), 2):
print(x[i])
4.
def remove_chars(x,y):
if y <len(x):
print(x[y:])
x = input("Give me a phrase: ")
y = int(input("Give me a length: "))
remove_chars(x,y)
5.
numbers_x = [10, 20, 30, 40, 10]
numbers_y = [75, 65, 35, 75, 30]
def first_and_last_the_same(list):
print("Given list:", list)
if list[0] == list[len(list)-1]:
print("Result is true")
else:
print("Result is false")

first_and_last_the_same(numbers_x)
first_and_last_the_same(numbers_y)
6.
list = [10, 20, 33, 46, 55]
print("Given list is ", list)
print("Divisible by 5")
for i in list:
if i%5 == 0:
print(i)
7.
z=0
x=0
str_x = "Emma is good developer. Emma is a writer"
try:
while x<len(str_x):

x = str_x.index("Emma",x)
x=x+1
z=z+1

except ValueError:
if z == 1:
print("Emma appeared",z ,"time")
else:
print("Emma appeared",z ,"times")
8.
x=0
y=" "
while x<5:
x=x+1
print((str(x)+y)*(x))
9.
# check if it is a number
def check_palindrome(x):
i=0
y=len(x)
while i0:
list.append(x % 10)
x = x//10

for i in list:
print(i, “”, end=””)
print(“”, end=”\n”)
12.
def cal_tax(income):
print(“Income: “, income)

if income <= 10000:
y=0
elif income <=20000:
y = 0.1*(income-10000)
else:
y = (1000+(income-20000)*0.2)
print("Tax:",'${:,.0f}'.format(y))
x = int(input("Give me your income:"))

cal_tax(x)

13.
x=1
while x <11:
y=1
while y 0:
print(“* “*x)
x=x-1
15.
x = int(input(“base: “))
y = int(input(“exponent: “))
def exponent(base, exp):
print(base, “raises to the power of”, exp,”is:”, base**exp,”i.e. (“, end=””)
for i in range (0, exp-1):
print(base, “*”, end=””)
i+=1
print(base,” = “,base**exp,”)”, sep=””)
exponent(x, y)

Thank you. it was fun 🙂

Reply
- sai says
  
  June 13, 2024 at 3:45 pm
  
  for exercise #9 :
  this is way better to understand:
  num=int(input(“enter the number :”))
  print(“original number”, num)
  num_str=str(num)
  if num_str==num_str[::-1]:
  print(“Yes.given number is palindrome number”)
  else:
  print(“No.given number is not palindrome number”)
  
  Reply
  - Sadegh says
    
    August 15, 2024 at 4:51 pm
    
    This is exactly what i did
    
    Reply
Deepak G says

February 13, 2024 at 1:10 pm

Exercise 11:

n = 7536
l= ” ”
for i in str(n):
l = ” “+i + l
print(l)

Reply
bhoomi says

February 7, 2024 at 7:25 pm

The alternate way that I found to do Ques. 9

num = str(input(“Please enter the number: “))
if num[: :] == num[len(num): :-1]:
print(“True”,num,”is a palindrome number”)
else:
print(“False”,num,”is not a palindrome number”)

Reply
Manpreet says

January 29, 2024 at 11:41 am

Exercise 6: Display numbers divisible by 5 from a list
l1=[10, 20, 33, 46, 55]
for num in l1:
if(num%5==0):
print(“divisible by 5:”,num)

Reply
Manpreet says

January 29, 2024 at 11:40 am

#Exercise 6: Display numbers divisible by 5 from a list
l1=[10, 20, 33, 46, 55]
for num in l1:
if(num%5==0):
print(“divisible by 5:”,num)

Reply
Manpreet says

January 29, 2024 at 11:20 am

#Exercise 5: Check if the first and last number of a list is the same
numbers_x = [10, 20, 30, 40, 10]
numbers_y = [75, 65, 35, 75, 30]
if(numbers_x[0]==numbers_x[-1]):
print(“result is true for number_x”)
else:
print(“result is false for number_x”)

if(numbers_y[0]==numbers_y[-1]):
print(“result is true for numbers_y”)
else:
print(“result is false for numbers_y”)

Reply
Manpreet says

January 29, 2024 at 10:41 am

exercise4:Remove first n characters from a string
str=”pynative”
str1=str[4:]
str2=str[2:]
print(str1,str2)

Reply
Manpreet says

January 29, 2024 at 10:28 am

exercise3:
#Print characters from a string that are present at an even index number
str=”pynative”
str1=str[0:7:2]
for i in str1:
print(i)

Reply
Al Foy says

January 22, 2024 at 1:08 am

This is cool, thank you!

Exercise 2.

—————— python code ——————————-
numb_now = 0
for each in range( 10 ):
numb_now = each
#numb_now = numb_now + 1
numb_prev = numb_now – 1
if numb_prev < 0:
numb_prev = 0
sum = numb_now + numb_prev
print( 'Current Number = ', numb_now, 'Previous Number = ', numb_prev, 'Sum:', sum )

———————- Output —————————-
Current Number = 0 Previous Number = 0 Sum: 0
Current Number = 1 Previous Number = 0 Sum: 1
Current Number = 2 Previous Number = 1 Sum: 3
Current Number = 3 Previous Number = 2 Sum: 5
Current Number = 4 Previous Number = 3 Sum: 7
Current Number = 5 Previous Number = 4 Sum: 9
Current Number = 6 Previous Number = 5 Sum: 11
Current Number = 7 Previous Number = 6 Sum: 13
Current Number = 8 Previous Number = 7 Sum: 15
Current Number = 9 Previous Number = 8 Sum: 17

Executed in: 0.02 sec(s)
Memory: 4236 kilobyte(s)

Reply
Osama Aftab says

January 14, 2024 at 11:05 pm

Exercise 9: Check Palindrome Number
num1= int(input(‘> ‘))
num2 = str(num1)[::-1]
if num1 == int(num2):
print(‘Yes. given number is palindrome number’)
else:
print(‘No. given number is not palindrome number’)

Reply
Dona Chaudhuri says

January 7, 2024 at 2:14 am

#Excercise 15:
def exponent(base,exp):
if exp == 0:
return 1
else:
print(“Result: “,base**exp)
base = int(input(“Enter base: “))
exp = int(input(“Enter exp: “))

exponent(base,exp)

Reply
- Nethra says
  
  February 17, 2024 at 9:20 am
  
  def exe_fifteen():
  base = int(input(“Enter the base num : “))
  exp = int(input(“Enter the exp num : “))
  print(pow(base,exp))
  exe_fifteen()
  
  Reply
Dona Chaudhuri says

January 6, 2024 at 12:52 am

Exercise 9: Check Palindrome Number

number = input(“Enter a 3 digit number: “)
print(“Original number “,number)
reverse_num = str(number[::-1])

if int(number) == int(reverse_num):
print(“Yes. given number is palindrome number”)
else:
print(“No. given number is not palindrome number”)

Reply
Pintu says

January 2, 2024 at 10:06 pm

## Print multiplication table from 1 to 10

def MultiTable(n):
for i in range(1, n+1):
for j in range(1, 11):
print(i*j, end=” “)
print(“”)

num = int(input(“Please Enter a value :”))
obj = MultiTable(num)

Reply
Talal says

November 19, 2023 at 11:10 am

Exercise 14: Print a downward Half-Pyramid Pattern of Star (asterisk)

for i in range(5, 0, -1):
print(“*” * i)

Reply
Talal says

November 19, 2023 at 11:03 am

list1 = [10, 20, 25, 30, 35]
list2 = [40, 45, 60, 75, 90]

odd = [x for x in list1 if x % 2 != 0]
even= [x for x in list2 if x % 2 == 0]

print(odd + even)

Reply
- Areena says
  
  March 16, 2024 at 10:23 pm
  
  Thanks🌸
  
  Reply
Talal says

November 19, 2023 at 10:56 am

def exponent(base, exp):
return base**exp
print(f'{5} raises to the power of {4} is: {exponent(5, 4)}’)

Reply
- Pintu says
  
  January 2, 2024 at 10:03 pm
  
  def Exponent(base, exp):
  value = base
  for i in range (1, exp) :
  value = value*base
  print(f”The value of {base} to the power {exp} is {value}”)
  print(value)
  
  b = int(input(“Please enter a int value : “))
  e = int(input(“Please enter a exponent value: “))
  
  obj = Exponent(b,e)
  
  Reply
ian says

November 16, 2023 at 4:55 am

1.

number1 = int(input(“Enter value x: “)
number2 = int(input(“Enter value y: “)

print(x * y) if x * y <= 1000 else print(x + y)

Reply
- Chelly says
  
  December 9, 2023 at 3:42 pm
  
  👍👍
  
  Reply
Clara M says

October 25, 2023 at 2:13 am

For Exercise 8, taking advantage of the properties of a string object:

def number_pyramid(number):

for i in range(1, number + 1):
print(str(i)*i)

Reply
LucasM says

October 5, 2023 at 3:11 pm

Exercise 5

from typing import List

def is_same(my_list: List[int]) -> bool:
print(f”Given list: {my_list}”)
return my_list[0] == my_list[-1]

def main():
numbers_x = [10, 20, 30, 40, 10]
numbers_y = [75, 65, 35, 75, 30]

print(is_same(numbers_x))
print(is_same(numbers_y))

if __name__ == “__main__”:
main()

Reply
RGBSupreme says

September 30, 2023 at 3:38 am

Exercise 15:

#Exponent value from user.
input_exp = input(“Enter a non negative integer for your exponent:”)
print(“”,input_exp)

#Base value from user.
input_base = input(“Enter any integer for your base:”)
print(“”,input_base)

#Input value conversions from string to integer.
exp = int(input_exp)
base = int(input_base)

print(” “)

#Result
print(“Base integer”, input_base , “to the power of base exp”, input_exp, “is equal to”,base ** exp,”.”)

Takes input from a user instead.

Reply
michalx says

September 8, 2023 at 1:04 am

Exercise 9: Check Palindrome Number

Pros: Readable, simple, and straightforward for anyone to understand.
Cons: Involves converting numbers to strings, which might not be as efficient for very large numbers.

def is_palindrome(number):
number_str = str(number)
palindrome = list(reversed(number_str))
print(f”Orginal number is: {number}”)
if list(number_str) == palindrome:
print(“Yes. the given number is palindrome number”)
else:
print(“No. The given number is not palindrome number”)

is_palindrome(125)

Reply
- Gol D. Roger says
  
  September 15, 2023 at 10:08 pm
  
  ori_num = input(‘input text or a number’)
  if ori_num == ori_num[::-1]:
  print(‘is a palindrome’)
  else:
  print(‘is not a palindrome’)
  
  I think this way is more easy
  
  Reply
  - Phindulo Ratshilumela says
    
    November 6, 2023 at 1:09 am
    
    yep agree
    
    Reply
michal says

September 8, 2023 at 1:02 am

Exercise 9: Check Palindrome Number

Pros: Readable, simple, and straightforward for anyone to understand.
Cons: Involves converting numbers to strings, which might not be as efficient for very large numbers.

def is_palindrome(number):
number_str = str(number)
palindrome = list(reversed(number_str))
print(f”Orginal number is: {number}”)
if list(number_str) == palindrome:
print(“Yes. the given number is palindrome number”)
else:
print(“No. The given number is not palindrome number”)

is_palindrome(125)

Reply
- vishnu pk says
  
  December 1, 2023 at 6:46 pm
  
  def paliandrome(number):
  print(“orginal number ” + str(number))
  if str(number)==str(number)[::-1]:
  print(“Yes. given number is palindrome number”)
  else:
  print(“No. given number is not palindrome number”)
  
  Reply
pullamma says

July 24, 2023 at 11:41 am

thanks

Reply
pullamma says

July 24, 2023 at 11:39 am

string = "/*Jon is @developer & musician!!" special_symbols = "!@#$%^&*()_+-={}[]|\:;\"',.?/~`" for char in special_symbols: string = string.replace(char, "#") print(string)

I don’t know this code

Reply
syb02 says

July 20, 2023 at 10:31 pm

Exercise 12:

income = 25000 tax = 0 while income > 10000: if income > 20000: tax += (income - 20000) * 20/100 income = 20000 elif income > 10000: tax += (income - 10000) * 10/100 income = 10000 print(tax)

Reply
- ffrancisco says
  
  August 16, 2023 at 2:37 pm
  
  #Newbie ME!
  #for salary above 21K
  
  x = 10000
  
  entry = input(“enter salary here: ” )
  
  deduct1 = ((float(entry) – 10000))
  deduct2 = ((float(deduct1) – 10000 ))
  
  first_tax = (x * .0)
  second_tax = (x * .1)
  remaining = ((int(deduct2)*.2))
  total_tax = ((float(first_tax) + float(second_tax) + float(remaining)))
  
  print (x,”x”,”0%”,”+”, x,”x”,”10%”,”+”,remaining,”x”,”25%”, “=”, total_tax)
  
  Reply
- Mehnaj Chowdhury says
  
  August 20, 2023 at 7:10 pm
  
  def tax_amount(income_amount):
  first=10000
  second=10000
  remains=income_amount-(first+second)
  tax=int(first*0.0+second*0.10+remains*0.20)
  print(tax)
  inp=int(input(“Enter the taxable income: “))
  tax_amount(inp)
  
  Reply
- jonjon says
  
  January 6, 2024 at 12:10 am
  
  the easy solution for me is from the formula itself:
  print(tax_payable)
  income = 45000
  x = income – 35000
  taxpayable1 = x * 10/100
  b = income – 20000
  taxpayable2 = b *20/100
  
  total_tax = taxpayable1+ taxpayable2
  print (total_tax)
  
  Reply
syb02 says

July 20, 2023 at 10:17 pm

Exercise 11:

num = 7536 reversed_num_str = str(num)[::-1] for c in reversed_num_str: print(c, end=" ")

Reply
syb02 says

July 20, 2023 at 9:32 pm

Exercise 8:

row_count = 9 for i in range(1, row_count + 1): print((str(i) + " ") * i)

Reply
syb02 says

July 20, 2023 at 9:12 pm

I think this is a more accurate solution for Exercise 7 -> Solution 2.

1- There is no need to go through the entire str_x (For example, if the substring you are looking for is 4 characters long, it is pointless to search for it in the 3rd-to-last index of str_x because the largest substring you can get from the 3rd-last index will be 3 characters long.)

2- There is no need to take all substrings of str_x and compare them. It is sufficient to look only for those whose first character matches the searched phrase’s first character. (I mean, it’s pointless to compare “good” with “Emma”)

str_x = "Emma is good developer. Emma is a writer" substr = "Emma" count = 0
for i in range(len(str_x) - (len(substr) - 1)): if str_x[i] == substr[0]: count += str_x[i : i + len(substr)] == substr
print(count)

Reply
- syb02 says
  
  July 20, 2023 at 9:21 pm
  
  str_x = "Emma is good developer. Emma is a writer" substr = "Emma" count = 0
  for i in range(len(str_x) - (len(substr) - 1)): if str_x[i] == substr[0]: count += str_x[i : i + len(substr)] == substr
  print(count)
  
  Reply
Amir says

July 12, 2023 at 5:59 pm

Exercise 15:

def exp(b,e): y = b for i in range(1,e): b *= y return b
print(exp(5,4))

Reply
- Ind01 says
  
  August 17, 2023 at 7:00 pm
  
  #newincoding
  we can do in simpliest way:
  
  def exponent(base,exp):
  new_int=base**exp
  return new_int
  
  print(exponent(2,5))
  
  Reply
sfssadf says

July 12, 2023 at 5:42 pm

Exo 2 has difference between input code and output printing

Reply
LNT says

June 2, 2023 at 3:45 am

EXERCISE 9 Simpler method:

def isPalindrome(number): if number[::-1] == number: return "This is a palindrome" else: return "This isnt a palindrome"

Reply
- Diwakar says
  
  September 7, 2023 at 12:26 pm
  
  You will get below error:
  
  —————————————————————————
  TypeError Traceback (most recent call last)
  Cell In[64], line 11
  7 else:
  9 return “This isnt a palindrome”
  —> 11 isPalindrome(545)
  
  Cell In[64], line 3, in isPalindrome(number)
  1 def isPalindrome(number):
  —-> 3 if number[::-1] == number:
  5 return “This is a palindrome”
  7 else:
  
  TypeError: ‘int’ object is not subscriptable
  
  Reply
Sankar says

May 25, 2023 at 11:37 am

EXERCISES 15:–

def exponent(base, exp): i = base j = exp if j > 0: print(f"{i} raises to the power of {j} is: {i**j} ") print()
exponent(2, 5)

Reply
- Donald says
  
  July 4, 2023 at 2:02 pm
  
  def expo(base,exponent): return base**exponent
  expo(2,5)
  
  Reply
  - Danilo says
    
    September 28, 2023 at 11:23 am
    
    def exponent(base, exp):
    print(“base =”, base)
    print(“exponent =”, exp)
    result = 1
    for i in range(exp):
    result *= base
    print(base, “raises to the power of”, exp, “is:”, result)
    exponent(2,5)
    
    Reply
Trotun says

May 19, 2023 at 11:03 am

I think these excercises are very funny. I have looked a lot for something like this. Thanks a lot.

Reply
Galambos says

May 2, 2023 at 2:48 am

Exercise 13 is weird. For some reason, the expected output shows the first column being separated from the second (maybe with tab? not sure), even though the rest of the columns are not aligned.
On the other hand, the provided solution aligns none of the columns, and, for some reason, inserts 2 tabs at the end of each line.

My solution:
for i in range(1, 11): for j in range(1, 11): # insert tab after every number to keep the columns aligned: print(f"{i * j}\t", end=" ") # empty print statement breaks the line just fine, no need to insert tabs here: print("")

Reply
- Galambos says
  
  May 2, 2023 at 2:55 am
  
  for i in range(1, 11): for j in range(1, 11): # insert tab after every number to keep the columns aligned: print(f"{i * j}\t", end=" ") # empty print statement breaks the line just fine, no need to insert tabs here: print("")
  
  Reply
shreeji patel says

February 27, 2023 at 8:44 pm

correction :

def exponent(base, exp): return base**exp
exponent(2, 5)

Reply
- 0xAH says
  
  March 2, 2023 at 12:10 am
  
  You did not count this remark:
  ‘Note here exp is a non-negative integer, and the base is an integer.’
  
  Reply
  - SAI PAVITRA VANGALA says
    
    April 11, 2023 at 4:51 pm
    
    Hi! Could you please explain me the question number 15 solution code
    
    Reply
  - what is this site? says
    
    June 12, 2023 at 7:50 pm
    
    ok…
    def exponent(base, exp): if exp>0: return base**exp else: return('exp needs to be positive integer')
    exponent(2, 5)
    
    still way easier than what he wrote
    
    Reply
- Vinay Vishwakarma says
  
  April 27, 2023 at 4:36 pm
  
  def aaa(base, exponent): output = 0 output = base ** exponent return output
  base = 5 exponent = 4 aaa(base, exponent)
  
  Reply
- ANKUSH Saral says
  
  May 27, 2023 at 9:20 pm
  
  It’s good actually.
  
  Reply
Med_Ka says

February 26, 2023 at 8:15 pm

# Exercise 7: Return the count of a given substring from a string

text = input("Enter the text: ") search_word = input("Enter the word to search: ")
text_lower = text.lower() search_word_lower = search_word.lower() count = text_lower.count(search_word_lower)
if count == 0: print("The word is not present in the text.") else: print(f"The word '{search_word}' appears {count} times in the text.")

Reply
Med_Ka says

February 26, 2023 at 6:45 pm

#Exercise 6: Display numbers divisible by 5 from a list

#Create a list as a global variable my_list=[] #Function to create a list def create_list(n): for i in range(n): value = int(input("Enter value {}: ".format(i+1))) my_list.append(value) return my_list #Function to divide a string by a number def div_by_number(m): for i in range(len(my_list)): if my_list[i] % m == 0: print(my_list[i])
#Main program #insert the number of elements for the list x = int(input("How many elements the list must contain ? ")) Given_list = create_list(x) print("Given list is : ", Given_list)
#enter the divisor : y = int(input("The elements of the list are divisible by : ")) #return the function by printing the elements which are divisible by the number you have chosen print("The elements of the Given list which are divisible by ",y ," are : ") div_by_number(y)

Reply
Med_Ka says

February 26, 2023 at 3:24 am

Exercise 5: Check if the first and last number of a list is the same

# Function to create a list def create_list(n): my_list = [] for i in range(n): value = input("Enter value {}: ".format(i+1)) my_list.append(value) return my_list
# Function to compare the first and the last element of a list def compare_first_last(lst): return lst[0] == lst[-1]
# Insert number of elements and the elements of the list x = int(input("Enter the number of elements: ")) This_List = create_list(x) print(This_List) Compare = compare_first_last(This_List) print(Compare)

Reply
Issam says

February 24, 2023 at 5:53 am

Ex 14
solution

for i in reversed(range(6)): for j in range(i): print('*',end=' ') print('\n')

Reply
steve k says

February 7, 2023 at 11:54 pm

I get what the second exercise is doing , new to posting comments or even reading tutorials on the web, but can someone fix the expected output to be what the code solution is or fix the code on exercise 2 to match output. Old QA guy learning python LOL. I iterated over a list instead of a range function but I assume if they change the range to 0,10 it would match expected output.

Reply
- Melvin J says
  
  March 21, 2023 at 5:05 am
  
  Hey Steve, this was my solution to match the output
  start_num = 10 print("Printing current and previous number sum in a range(%d)" % (start_num)) output_txt = "Current Number %d Previous Number %d Sum: %d"
  for count in range(start_num): curr_num = count prev_num = (curr_num - 1) if prev_num < 0: prev_num = 0 total_sum = (curr_num + prev_num)
  print(output_txt % (curr_num,prev_num,total_sum))
  
  Reply
Iván says

February 4, 2023 at 5:06 am

Is this solution right for exercise 8?

n = 1 while n <=5: print(" ".join((str(n)*n))) n+=1

Reply
Krishna Shyam says

January 5, 2023 at 10:30 am

#Write a program to extract each digit from an integer in the reverse order def extract(number): reverse _ list = [] for i in number[::-1]: reverse _list = i print(reverse_ list, end = " ") extract(input("enter the digits >> "))
Is this the right order for the given question?

Reply
- akash says
  
  January 16, 2023 at 1:21 am
  
  def rev_int(given_int): new_int = "" for i in range(1, len(given_int)+1): new_int += given_int[-i] new_int += " " return new_int
  print(rev_int("8687"))
  
  Reply
  - akash says
    
    January 16, 2023 at 1:22 am
    
    def rev_int(given_int): new_int = “” for i in range(1, len(given_int)+1): new_int += given_int[-i] new_int += ” ” return new_int
    print(rev_int(“8687”))
    
    Reply
    - akash says
      
      January 16, 2023 at 1:24 am
      
      def rev_int(given_int): new_int = "" for i in range(1, len(given_int)+1): new_int += given_int[-i] new_int += " " return new_int
      print(rev_int("8687"))
      
      Reply
  - Pooky says
    
    February 15, 2023 at 2:36 am
    
    Its way easier If you do that in this way:
    
    stw = (input("Enter your Integer: ")) revstr = stw[::-1] for i in revstr: revint = int(i) print(i, end=" ")
    
    Reply
- manish kumar says
  
  January 17, 2023 at 5:22 pm
  
  x=input("Enter the integer") for i in x: print(x[int(len(x))-int(i)])
  
  Reply
- Shahzad Ahmad says
  
  February 1, 2023 at 5:41 pm
  
  def reverse_digit(num): print('Digit in reverse order: ', end=" ") while num > 0: remainder = num % 10 num = num // 10 print(remainder, end=' ')
  num = int(input('Number: ')) reverse_digit(num)
  
  Reply
  - Arca says
    
    February 27, 2023 at 12:20 am
    
    a=str(input("original number ")) for i in range(len(a)): if a[i] == a[-(i+1)]: reponse=("Yes. given number is palindrome number") else: reponse=("No. given number is not palindrome number") print(reponse)
    
    Reply
Harlan Silva says

October 27, 2022 at 7:06 am

Hi, people!

I made exercise 9 differently compared with most of the examples here. It’s not in English, but I think it’s possible to understand the logic. Is it worst than the suggested in the solutions?

# Criar função para identificar palíndromo def identificar_palindromo (num: str) -> bool: # Criar identificação do primeiro e terceiro números num_1 = int(num[0]) num_3 = int(num[2]) if num_1 == num_3: resultado = True print(f"O número escolhido é um palíndromo.\nO primeiro é {num_1} e o terceiro é {num_3}") else: resultado = False print(f"O número escolhido não é um palíndromo.\nO primeiro é {num_1} e o terceiro é {num_3}") return resultado
# Chamar a função através da variável num_palindromo = identificar_palindromo(input("Digite um número: "))

Reply
- Definitelynot200 says
  
  November 12, 2022 at 8:30 am
  
  I think it’s not flexible because you’re just comparing the index 0 and index 2 of the given input. If I just put a 4 digit input with a value of 4144 or a 5 digit input 53235 for example, it will throw true in the 4144 and false in the 53235 which is not. That’s what I think.
  
  Reply
- sajib ahamed says
  
  November 16, 2022 at 1:35 am
  
  def cheack_palindrome(number): print('Original number ',number) if str(number) ==str(number)[::-1]: print('Yes. given number is palindrome number') else: print('No. given number is not palindrome number.')
  cheack_palindrome(121) cheack_palindrome(125)
  
  Reply
Zsolt Remenyi says

October 25, 2022 at 9:33 pm

Hi,

Love the site. I think ‘Exercise No. 13’ would look better aligned.

for x in range(1, 11): for y in range(1, 11): print(f"{x*y:>4}", end = " ") print("\n")

Reply
Mustafa says

September 26, 2022 at 4:54 pm
I finish like this😅
```
Exercise 10: Create a new list from a two list using the following condition

list1 = [10, 20, 25, 30, 35]
list2 = [40, 45, 60, 75, 90]
list3 = []

for i in list1:
    if i % 2 == 1:
        list3.append(i)
    else:
        continue

for j in list2:
    if j % 10 == 0:
        list3.append(j)
    else:
        continue

print(list3)
```
Reply
- Jig says
  
  October 9, 2022 at 7:02 pm
```
#This will run quick and save your memory
l1 = [10, 20, 25, 30, 35]
l2 = [40, 45, 60, 75, 90]
l3=[]
for i in l1:
    l3.append(i) if i%2!=0 else l3
for i in l2:
    l3.append(i) if i%2==0 else l3
print(l3)
```
  Reply
- pyboy says
  
  October 14, 2022 at 9:47 pm
  
  ex:15
  
  def exponent(base, exp): print("le numero introduit est ",base, "donc le resultat est", base**exp)
  numeri=exponent(float(input("mettre la base\n")),float(input("mettre l'exp\n")))
  
  Reply
  - Aman says
    
    October 15, 2022 at 2:41 pm
    
    how about this for a solution to question 10
    
    list1 = [10, 20, 25, 30, 35] list2 = [40, 45, 60, 75, 90] newlist1 = [i for i in list1 if i%2!=0] newlist2 = [i for i in list2 if i%2==0] newlist = newlist1 + newlist2 print(newlist)
    
    Reply

lily fullery says

September 24, 2022 at 9:12 pm

#print the sum of  current and previous number 
def current_previous():
    cnum = 0
    pnum = 0
    for i in range (10):
        if(i == 0):
            print("currentnumber", 0 ,"and"," previousnum ", 0 , "sum :", 0)
        else:
            cnum = 0 + i 
            pnum = cnum-1
            allover = cnum + pnum
            print("currentnumber", cnum ,"and"," previousnum ", pnum , "sum :", allover)
current_previous()

#string from the user and display character that present at an even number 

user = input("enter any word :")
def evenstring(user):
    for i in range(len(user)):
        if(user.index(user[i])%2 == 0):
            print(user[i])
evenstring(user)

—————————————————————————-

#remove first n characters from a string 
def first_n(word , n):
    word2 = word[n: ]
    return word2
word = 'pynative'
n = 1
value = first_n(word , n)
print(value)

——————————————————————————

#Printing stars
#print the pattern
n = 6
a = None
for i in range(1,n):
    a = (str(i)+" ") * i
    print(a)

———————————————————————————-

#  create a new list from a two list using the following condition
list1 = [10, 20, 25, 30, 35]
list2 = [40, 45, 60, 75, 90]
mylist = []
for i in list1:
    if(i%2 == 1):
        mylist.append(i)
for i in list2:
    if(i%2 == 0):
        mylist.append(i)

print("result list : ",mylist)

————————————————————————————

#Write a Program to extract each digit from an integer in the reverse order.
num = 7536
strnumber = str(num)
for i in range(1,len(strnumber)+1):
    a = i-(i+i)
    print(strnumber[a], end='')

———————————————————————————-

##double nested loop
for i in range(1,11):
    a = 1
    for i in range(1,11):
        b = a * i
        a = a +1
        print(b, end =' ')

——————————————————————————–

# Print downward Half-Pyramid Pattern with Star
a = ""
n = 5
t = n
for i in range(n):
    a = '* ' * t 
    t = t - 1
    print(a)

meera says

June 8, 2023 at 4:10 pm

you replied on my birthday

Reply

vini1955 says

September 23, 2022 at 3:09 pm

very helpful.

Reply
- Vishal says
  
  September 26, 2022 at 9:49 am
  
  I am glad it helped you.
  
  Reply
Eli says

September 8, 2022 at 5:47 am
My solution for exercise 2:
```
lst = range(10)
previous = 0

for value in lst:
    previous = max(value - 1, 0)
    print(f'Current: {value} | Previous: {previous} | Sum: {value + previous}')
```
Not sure how that compares functionally with teach’s code
Reply
- sajib ahamed says
  
  November 16, 2022 at 1:38 am
  
  previous_number = 0 for i in range(0,10): current_number = i print('Current Number %d Previous Number %d Sum: %d'%(current_number,previous_number,(current_number+previous_number))) previous_number = i
  
  Reply
Eli says

September 8, 2022 at 5:43 am

Vishal, this is great. Thank you so much. It’s exactly what I needed.

I think that you might have an error in the solution for Ex. 2.
In the for loop, you define x_sum, but in the print statement, you reference previous_num + i, which is the left side of the equation. I think you meant to reference x_sum there. If I’m wrong about that, please let me know.

It confused me at first, which was a good thing because I had to figure things out for myself.

Reply
- Vishal says
  
  September 26, 2022 at 9:56 am
  
  Hi Eli, the x_sum and previous_num + i produce the same result. But for simplicity, i have replaced x_sum with previous_num + i
  
  Reply

Hardik Saggar says

September 4, 2022 at 3:56 pm

# Exercise 15
base = 2
exp = 4
print(f"{base} raises to the power of {exp}: ",base**exp)

Taimoor AbdULLAH says

August 17, 2022 at 1:36 pm

# Exercise # 15

base = int(input("Base: "))
exponent = int(input("Exponent: "))
po = str(base ** exponent)
exp = ((str(base) + " * ") * exponent).strip()
print(str(base), "raised to the power of", str(exponent), ":", po, "\ni.e (", exp[:len(exp) - 2], ") = ", po)

deus says

September 2, 2022 at 3:03 pm

def exponent(base,power):
    x=base**power
    print(base, "raises to the power of", power, "is: ", x)
exponent(5,2)

valid??

Hardik Saggar says

September 4, 2022 at 3:59 pm
```
exponent(2,5) which will give you 32 in output
but here is a simpler way
base = 2
exp = 4
print(f"{base} raises to the power of {exp}: ",base**exp)
```
in this you can increase its functionality by taking input from user and code would of lesser lines
Reply

Taimoor AbdULLAH says

August 17, 2022 at 12:57 pm
Here is my 2 line solution for exercise # 14
```
# Exercise # 14

for i in range(5, 0, -1):
    print("* " * i)
```
Reply
- Chinuogu David Enweani says
  
  August 24, 2022 at 10:46 pm
  
  You a goat bruh
  
  Reply
  - noname says
    
    November 15, 2023 at 8:35 pm
    
    try contributing then
    
    Reply
- Mehrdad says
  
  September 8, 2022 at 11:25 pm
```
i = 50
print('')
while i > 0 :
    print('* '*i)
    i -= 1
```
  Reply

Taimoor AbdULLAH says

August 17, 2022 at 12:30 pm

# Exercise 12
i = int(input("Enter income: "))
tax = 0
if i <= 1000:
    tax += 0
elif i <= 2000:
    tax += 1000
else:
    tax += (i - 20000) * .2 + 1000
print("Tax Collected:", tax)

Taimoor AbdULLAH says

August 17, 2022 at 12:08 pm
below is My 2 lines solution for exercise # 11
```
i = str(input("Enter an integer:"))
print("Reverse Integer is:", i[::-1])
```
Reply

Taimoor AbdULLAH says

August 17, 2022 at 12:04 pm

result = []
for i in range(2):
    print("Enter List ", i + 1)
    for j in range(5):
        inp = int(input())
        if inp % 2 != 0 and i == 0:
            result.append(inp)
        elif inp % 2 == 0 and i == 1:
            result.append(inp)
print("Result List", result)

Taimoor AbdULLAH says

August 17, 2022 at 12:09 pm

Exercise # 10 with user input list

Reply

Shaun says

August 17, 2022 at 2:27 am
I was proud to get exercise 13 in one line. I don’t know if it’s the most efficient, but it is just one long line of code.
```
def exponentExplained(base, exponent):
    print('{0} raised to the power of {1} : {2} i.e. ({3})'.format(base, exponent, base**exponent,'*'.join([x for x in [str(base)]*exponent])))

exponentExplained(2, 5)
```
Reply
- Shaun says
  
  August 17, 2022 at 2:29 am
  
  Sorry. This was exercise 15.
  
  Reply

Anon says

August 15, 2022 at 4:19 pm

Here is my solution to exercise 9, did it for any palindromes. Works with numbers too.


word = input('was da word palidrom??: ')
for check in range(len(word)):
    if word[check] == word[len(word)-check-1]:
        continue
    else:
        print('no')
        quit()
print('yes')

Sri says

August 25, 2022 at 11:05 am

Hello! Thanks for the insight.
I am a beginner. I did the palindromes exercise like this

a = str(input("Please enter the text: "))
if a[:len(a)] == a[::-1]:
    print ("Palindrome")
else:
    print ("Not a Palindrome")

Sophie says

August 10, 2022 at 10:46 am
This is my alternative solution for Exercise 11. I’m still a beginner in Python 😀
```
 number = 7536
print("Given number:", number)

str_x = str(number)

for num in str_x[::-1]:
    print(num, end=" ")
```
Reply
- Matt says
  
  November 17, 2022 at 1:11 pm
  
  a = 7536 for i in range(len(str((a)))-1,-1,-1): print(str(a)[i],end=" ")
  
  Reply

Sophie says

August 10, 2022 at 10:41 am

This is my alternative solution for Exercise 9. I understood it better this way 🙂

 def palindrome(number):
    print("Original number:", number)
    
    str_x = str(number) 
    original_num = str_x
    reversed_num = str_x[::-1]
    
    if original_num == reversed_num:
        print("Given number is a palindrome,")
        
    else:
        print("Given number is not a palindrome.")

Sahadad HT says

August 1, 2022 at 8:45 pm

I’m a beginner in python. I have tried exercise no 13 like this


print("Table of 1-10 \t")
for i in range(1,11):
    for j in range(1,11):
        print(i,"*",j,"=",i*j,end="\t\t")
    print("\t")

result

1 * 1 = 1		1 * 2 = 2		1 * 3 = 3		1 * 4 = 4		1 * 5 = 5		1 * 6 = 6		1 * 7 = 7		1 * 8 = 8		1 * 9 = 9		1 * 10 = 10			
2 * 1 = 2		2 * 2 = 4		2 * 3 = 6		2 * 4 = 8		2 * 5 = 10		2 * 6 = 12		2 * 7 = 14		2 * 8 = 16		2 * 9 = 18		2 * 10 = 20			
3 * 1 = 3		3 * 2 = 6		3 * 3 = 9		3 * 4 = 12		3 * 5 = 15		3 * 6 = 18		3 * 7 = 21		3 * 8 = 24		3 * 9 = 27		3 * 10 = 30			
4 * 1 = 4		4 * 2 = 8		4 * 3 = 12		4 * 4 = 16		4 * 5 = 20		4 * 6 = 24		4 * 7 = 28		4 * 8 = 32		4 * 9 = 36		4 * 10 = 40			
5 * 1 = 5		5 * 2 = 10		5 * 3 = 15		5 * 4 = 20		5 * 5 = 25		5 * 6 = 30		5 * 7 = 35		5 * 8 = 40		5 * 9 = 45		5 * 10 = 50			
6 * 1 = 6		6 * 2 = 12		6 * 3 = 18		6 * 4 = 24		6 * 5 = 30		6 * 6 = 36		6 * 7 = 42		6 * 8 = 48		6 * 9 = 54		6 * 10 = 60			
7 * 1 = 7		7 * 2 = 14		7 * 3 = 21		7 * 4 = 28		7 * 5 = 35		7 * 6 = 42		7 * 7 = 49		7 * 8 = 56		7 * 9 = 63		7 * 10 = 70			
8 * 1 = 8		8 * 2 = 16		8 * 3 = 24		8 * 4 = 32		8 * 5 = 40		8 * 6 = 48		8 * 7 = 56		8 * 8 = 64		8 * 9 = 72		8 * 10 = 80			
9 * 1 = 9		9 * 2 = 18		9 * 3 = 27		9 * 4 = 36		9 * 5 = 45		9 * 6 = 54		9 * 7 = 63		9 * 8 = 72		9 * 9 = 81		9 * 10 = 90			
10 * 1 = 10		10 * 2 = 20		10 * 3 = 30		10 * 4 = 40		10 * 5 = 50		10 * 6 = 60		10 * 7 = 70		10 * 8 = 80		10 * 9 = 90		10 * 10 = 100

Ahmed says

July 31, 2022 at 1:54 pm

I try to solve exercise 13 in a different way which output looks good


for i in range(1,11):
    for j in range(1,11):
        n = i*j
        if n // 10 == 0:
            print(n,end=' '*2)         
        else: 
            print(n,end=' ')
        
    print()

the result

1  2  3  4  5  6  7  8  9  10 
2  4  6  8  10 12 14 16 18 20 
3  6  9  12 15 18 21 24 27 30 
4  8  12 16 20 24 28 32 36 40 
5  10 15 20 25 30 35 40 45 50 
6  12 18 24 30 36 42 48 54 60 
7  14 21 28 35 42 49 56 63 70 
8  16 24 32 40 48 56 64 72 80 
9  18 27 36 45 54 63 72 81 90 
10 20 30 40 50 60 70 80 90 100

Sri says

August 25, 2022 at 11:10 am

Hello. Thanks for the insight.

I did exercise 13 like this.

x = int(input("What is the number : "))
y = int(input("Upto which number : "))
a = range(1,x+1)
b = range(1,y+1)
for i in a:
    for v in b:
        lst = (i*v)
        frmt = f"{lst:4.0f}"
        print(frmt,end="")
    print("")

harshul Patel says

November 8, 2022 at 5:58 pm

its look really good . Thank You..:)

Reply
apers says

June 10, 2024 at 11:17 pm

“””
Exercise 14: Print a downward Half-Pyramid Pattern of Star (asterisk)
* * * * *
* * * *
* * *
* *
*
“””

for i in range(5, 0, -1):
print(str(‘*’) * i)

Reply

luke says

July 24, 2022 at 2:28 pm

my solution for exercise number 6:

listnum = [5,25,55,15544,33,1355]
def fivenum(list):
    for x in list:
        c= x%5
        if c== 0:
            print(x)

        else:
            continue
fivenum(listnum)

Zane Hewgley says

July 13, 2022 at 3:58 pm
As with a lot of people, I am very new to this.
On question 1, I got the correct output but wrote it completely different than the solution.
Could someone explain to me why it is recommended to write it out the solution way over my answer..
```
number1 = 20
number2 = 30

sum = number1+number2
product = number1*number2
if product  1000:
    print("The result is " + str(sum))
```
Reply
- Zane Hewgley says
  
  July 13, 2022 at 3:59 pm
```
number1 = 20
number2 = 30

sum = number1+number2
product = number1*number2
if product  1000:
    print("The result is " + str(sum))
```
  Reply
  - Zane Hewgley says
    
    July 13, 2022 at 4:01 pm
    for some reason its not printing the last part.
    
    elif product > 1000: print("The result is " + str(sum))
    Reply
    - Zane Hewgley says
      
      July 13, 2022 at 4:05 pm
      
      sorry for the multiple comments…
      I realize it is leaving out stuff.
      it says-
      if product 1000:
      print("The result is " + str(sum))
      
      Reply
      - D Whitt says
        
        July 28, 2022 at 8:50 pm
        
        I’m not sure if I’m reading it right or if something is missing. But your code says:
        
        if product 1000:
        
        I’m guessing you meant to put
        
        if product is > 1000: print("The result is " + str(sum))
        
        This says if you multiply and the answer is over 1000 the print the sum instead.
        and instead of elif, you can do "else" which should cover anything being <= 1000
Sreehari m n says

July 11, 2022 at 1:17 pm

Anybody have another solution for question 13 for multiplication table with only increment of j in nested loop?

Reply
- Ben says
  
  July 12, 2022 at 12:32 am
  Like this?
```
for j in range(1,11) :
    prev = j
    for j in range(1,11) :
        print(j*prev, end=" ")
    print("")
```
  Reply
  - Sreehari says
    
    July 12, 2022 at 12:33 pm
    
    no.for ex:
    for table of 2: 2, 2+2, 4+2, 6+2 like that. every number in a table will be incremented by the same number.
    
    Reply
    - Malrey says
      
      July 18, 2022 at 10:45 pm
      
      maybe like this? :
      
      for i in range (1,11) : number = 0 for j in range(1,11) : number += i print(number, end=' ') print("\t\t")
      
      Reply
- Sahadad HT says
  
  August 1, 2022 at 8:48 pm
  you can try this
```
print("Table of 1-10 \t")
for i in range(1,11):
    for j in range(1,11):
        print(i,"*",j,"=",i*j,end="\t\t")
    print("\t")
```
  Reply

Ben says

July 9, 2022 at 11:42 pm

One Liners (I’m a beginner but really enjoying the one-line problems)

number1 = int(input("Number1: ")); number2 = int(input("Number2: ")); sum = number1*number2; print(sum) if sum >= 1000 else print(number1+number2) #Exercise 1
low = int(input("Input Starting Value: ")); [ print("Current Number %i Previous Number %i Sum: %i" % (i, 0 if i == low else i-1, low if i == low else i+i-1)) for i in range(low, low+10) ] #Exercise 2
word = input("Enter String: "); strlen = len(word); print("Original String is %s\nPrinting only even index chars" % word);[ print(word[i]) for i in range(0,strlen,2) ] #Exercise 3
word = input("Enter String: "); remove = int(input("Enter Number of Chars to Remove: ")); print(word[remove:]) #Exercise 4
num_list = [10, 20, 30, 40, 10]; print("Given list %ls\nresult is %s" % (num_list, "True" if num_list[0] == num_list[-1] else "False")) #Exercise 5
num_list = [10, 20, 33, 46, 55]; print("Given list is", num_list, "\nDivisible by 5"); [ print(value) if value % 5 == 0 else print("",end="") for index, value in enumerate(num_list) ] #Exercise 6
input = "Emma is good developer. Emma is a writer"; print(input.count("Emma")) #Exercise 7
[ print(("%i ")%(i)*i) for i in range(1,6) ]  #Exercise 8
user_input = str(input("Enter String: ")); print("Is a valid Palindrome") if user_input == user_input[::-1] else print("Is not a valid Palindrome") #Exercise 9
list1 = [10, 20, 25, 30, 35]; list2 = [40, 45, 60, 75, 90]; list3 = []; [ list3.append(list1[index]) if int(list1[index]) % 2 != 0 else print("",end="") for index in range(len(list1)) ]; [ list3.append(list2[index]) if int(list2[index]) % 2 == 0 else print("",end="") for index in range(len(list2)) ]; print(list3) #Exercise 10
user_input = str(input("Input Integer: ")); print(user_input[::-1]); #Exercise 11
[ print(' '.join([str(i*j) for i in range(1,11)])) for j in range(1,11) ] #Exercise 13
[ print('* '*i) for i in range(5,0,-1)] #Exercise 14
base = int(input("Enter Base Value: ")); exponent = int(input("Enter Exponent Value: ")); product = base ** exponent; [print(f'{base} raised to the power of {exponent} is {product}')] #Exercise 1

Rajat Singh says

June 18, 2022 at 12:30 am
Exercise 15: Write a function called exponent(base, exp) that returns an int value of base raises to the power of exp.

#code:
```
def exponent(base,ex):
    exponen = base ** ex
    return exponen

A = exponent(5,4)
print(A)
```
Reply
- John says
  
  July 5, 2022 at 7:29 pm
  
  Please I cannot understand the palindrome answer. I need explaination
  
  Reply
- Dario says
  
  July 11, 2022 at 6:38 pm
  that´s correct, but you need to add a condition where it only makes the math when the exp is greater than 0 like it says on the problem statement
  I did it like this
```
def exponent(base, exp):
    while exp > 0:
        result = base ** exp
        return result

base = 2
exp = 5

print("{} raises to the power of {}:".format(base, exp), exponent(base, exp))
```
  Reply
Chellammal says

June 7, 2022 at 5:00 pm
One line solutions to excercises 8,13 and 14 based on list comprehension.
```
[ print(((str(i)+" ")*i).rstrip()) for i in range(1,6) ] # excercise 8
[ print(" ".join([str(j) for j in range(i,((i*10)+1),i)])) for i in range(1,11) ] #excercise 13
[ print((('*'+" ")*i).rstrip()) for i in range(5,0,-1) ] #excercise 14
```
Reply
- Pravin says
  
  June 23, 2022 at 10:55 am
  
  Why you used rstrip() in #exercise 8?
  Sorry, I’m new to Python and wondering about this usage on that code. Thank you!
  
  Reply
  - CODHOO says
    
    July 4, 2022 at 7:04 pm
    
    He forgot it was supposed to be for beginners
    
    Reply
- Ben says
  
  July 9, 2022 at 8:45 pm
  I was trying to wrap my head around how the bracketed lines work, and I think this is a little clearer to read. Really glad I saw your comment, though, this seems like really useful stuff!
```
 [ print(' '.join([str(i*j) for i in range(1,11)])) for j in range(1,11) ]  #exercise 13 
```
  Reply
  - Ahmed says
    
    July 31, 2022 at 12:27 pm
    I do it with one for loop
    
    num = 0 for i in range(1,6): num += 10 ** (i-1) x = list(str(num*i)) print(' '.join(x))
    Reply

Chellammal says

June 7, 2022 at 4:59 pm

One-line solutions to exercises 8,13 and 14 based on list comprehension.

[ print(((str(i)+" ")*i).rstrip()) for i in range(1,6) ] # excercise 8
[ print(" ".join([str(j) for j in range(i,((i*10)+1),i)])) for i in range(1,11) ] #excercise 13
[ print((('*'+" ")*i).rstrip()) for i in range(5,0,-1) ] #excercise 14

Hero Zero says

June 1, 2022 at 2:06 pm

for exercise no. 15 with required output:


def exponent(base, exp):
    result = base ** exp
    print(f'{base} raises to the power of {exp}: {result} i.e. ({base}', end="")
    for i in range(1, exp):
        print(f' * {base}', end="")
    else:
        print(f' = {result})')


exponent(2, 5)

output:

C:\Users\BG-PC164a\PycharmProjects\pythonProject\venv\Scripts\python.exe C:/Users/BG-PC164a/PycharmProjects/pythonProject/main.py
2 raises to the power of 5: 32 i.e. (2 * 2 * 2 * 2 * 2 = 32)

Process finished with exit code 0

Hero Zero says

June 1, 2022 at 2:13 pm


def exponent(base, exp):
    result = base ** exp
    print(f'{base} raises to the power of {exp}: {result} i.e. ({base}', end="")
    for i in range(1, exp):
        print(f' * {base}', end="")
    else:
        print(f' = {result})')


exponent(5, 2)

C:\Users\BG-PC164a\PycharmProjects\pythonProject\venv\Scripts\python.exe C:/Users/BG-PC164a/PycharmProjects/pythonProject/main.py
5 raises to the power of 2: 25 i.e. (5 * 5 = 25)

Process finished with exit code 0

Ash says

May 30, 2022 at 2:45 am
Using if for Exercise 3: Print characters from a string that are present at an even index number
```
 a=input('Write a word:-')

l=len(a)


for n in range(l):
    if n%2==0: # using if 
        print(a[n])
```
Reply

Tam says

May 21, 2022 at 4:32 pm

Answer to exercise 9: Check if the string is a palindrome

 I know this is more convenient answer:
text = input("Enter string ")
a= text[::-1]
if a==text:
    print("This is palindrome")
else: print("This is not palindrome")

But this is also good solution to understand the logic:

str1 = input("Enter string")
revlist = []
last_index = -1
for i in str1:
    revlist.append(str1[last_index])
    last_index -=1
    str2 = ""
    for i in revlist:
        str2+=i
if str1==str2:
    print("This is palindrom")
else:
    print("this is not palindrom")

aravindhan says

May 11, 2022 at 9:08 pm

solution to exercise 15:

def exponentiation(base,exp):
    result = (base ** exp)
    print(exp, " is the exponentiation of ", base)
    print("The result is: ", result)

exponentiation(2,5)

Techo says

April 29, 2022 at 8:33 am

A good place to start with problem-solving. Before I would read articles, and watch videos; but, after solving these problems I realized I had been wasting my precious time. Spent all night on them.
Thank you very much!

Reply
carisa says

April 5, 2022 at 2:46 am

This site is extreamly usfull, totaly recomend.

Reply
- Hendi says
  
  April 25, 2022 at 3:19 pm
  
  There is a problem in question number one, the expected output is different from the question (contradicted). Anyway thanks for your sharing.
  
  Reply
  - ashwin says
    
    May 1, 2022 at 12:01 pm
    
    yes right
    
    Reply
carisa says

April 5, 2022 at 2:44 am

this site is extreamly usfull,totaly recomend

Reply
deschreiber says

March 24, 2022 at 6:50 pm
Exercise 9. Palindrome numbers. Is it cheating to check for a palindrome number by converting the number to a string, checking with indexing, then converting it back to an int? It’s a much simpler solution.
```
num = 123321
num_str = str(num)
if num_str == num_str[::-1]:
    print('Yes, given number is a palindrome number.')
else:
    print('No, given number is not a palidrome number.')
```
Reply
- deschreiber says
  
  March 24, 2022 at 6:54 pm
  
  Actually, of course, there’s no need to convert the string back to a number.
  
  Reply
- saurav says
  
  April 7, 2022 at 12:05 pm
  
  what if num is given user and the user give ‘mom’ as input.
  
  Reply
  - Mashood Iqbal Khan says
    
    April 23, 2022 at 1:01 pm
    x=input("Check Palindrome Number : ") if x==x[::-1]: print("Original Number: {0} \nYes. Given Number is palindrome number.".format(x)) else: print("Original Number: {0} \nNo. Given Number is not palindrome number.".format(x))
    Reply

ABD RRAHIM says

March 14, 2022 at 3:31 pm

Exercise 11 :


a = 7536
a = str(a)
for i in range(len(a)-1,-1,-1):
    print(a[i],end=' ')

ABD RRAHIM says

March 14, 2022 at 3:34 pm


def renum(num):
  num = str(num)
  for i in range(len(num)-1,-1,-1):
    print(num[i],end=' ')
renum(7536)

sunil says

August 26, 2022 at 4:29 pm

n = 7536
reversed = int(str(n)[::-1])
print(reversed)

ABD RRAHIM says

March 14, 2022 at 3:08 pm

EX 15:


def power(base,exponent):
  x = 1
  a = 1
  while x<= exponent:
   for x in range(1,6) :
    a = a*base
    x+=1
  print(base, "raises to the power of", exponent, "is: ", a)
power(2,5)

wayne says

April 4, 2022 at 1:39 am

def exponent(base, exp):
    calculate = base**exp
    return calculate

base = int(input('Enter the base: '))
exp = int(input('Enter the exponent: '))

print('The answer is', exponent(base, exp))

foaad says

March 8, 2022 at 1:45 am
ex14:
```
for i in range(1,6):
   print("* "*(6-i))
```
Reply
- Ram says
  
  March 29, 2022 at 8:15 am
```
for i in range(10,0,-1):
    print('* ' * i)
```
  Reply
  - ashwin chavhan says
    
    May 5, 2022 at 7:27 pm
    i=5 while i<=5: print(i*("*") i=i+1
    
    is that ok?//
    Reply

Hamza Younsi says

February 20, 2022 at 4:34 pm

Exercise 10:

list1 = [10, 20, 25, 30, 35]
list2 = [40, 45, 60, 75, 90]
c = list()
g = list()
for i in range(0, 5):
    if list1[i] % 2 != 0:
        c = list1[i]
        g.insert(i, c)
line = len(g)
cal = len(list2)
for m in range(line, cal - 1):
    if list2[i] % 2 == 0:
        c = list2[i]
        g.insert(i, c)
print("result list : ",g)

DebTahir says

February 28, 2022 at 12:48 am

def CreateNewList(list1,list2):
    newList = []
    for i in list1:
        if(i%2) != 0:
            newList.append(i)
    for j in list2:
        if(j%2) == 0:
            newList.append(j)
    return newList

list1 = [10, 20, 25, 30, 35]

list2 = [40, 45, 60, 75, 90]
print(CreateNewList(list1,list2))
[25, 35, 40, 60, 90]

Laurits says

January 31, 2022 at 7:59 pm

Question number 15:

a = int(input("Enter the base: "))
b = int(input("Enter your ex: "))
print ("Your answer is: ", a**b)

Will do the same just way easier

Mohammed Faizan says

January 26, 2022 at 10:48 pm

EXCERCISE 15: I think we can directly calculate the power of using (**) function

def exponent(base,exp):
    return(base**exp)

exponent(5,4)

tg says

March 26, 2022 at 5:10 pm

Yes, exactly. I really don’t get why using loops (while, for, etc.. here).


def exponent(base, exp):
	print(f"base = {base} \nexponent = {exp}\n")
	print(f"{base} raises to the power {exp} : {base**exp} i.e. ({(exp-1)*(str(base)+' * ')+ str(base)} = {base**exp}) ")

Leon Reaplust says

January 23, 2022 at 2:16 pm

I did Exercise 9: Check Palindrome Number diffrently, using string for reversing number. It goes like this:

 x = int(input('Write number you think is palindrome: '))
rev_x= str(x)[::-1]

if str(x)==rev_x:
  print('Number ',x,' is palindrome')
else:
  print('Number ',x,' is not palindrome')

Leon Reaplust says

January 22, 2022 at 5:24 pm

Exercise 6: Display numbers divisible by 5 from a list soo list is inputed by user, hope you like it 🙂

 def div_by_5(l):
  for i in l:
    if i%5==0:
      print('Number ',i,' is divisible with 5')
    else:
      print('Number ',i,' is not divisilbe with 5')

lst = []
n = int(input('Write number of elements for your list: '))

for i in range(0,n):
  ele=int(input('Write number on your list: '))
  lst.append(ele)

print('Your list is: ',lst)
print(div_by_5(lst))

Leon Reaplust says

January 22, 2022 at 4:05 pm

s = str(input('Write your word: '))
n = int(input('Write number of letters that you want to remove from word: '))

if n < len(s):
  print('Word without first ',n,'letters is: ',s[n:])
else:
  print('n is too big!')

This is my solution for Exercise 4: Remove first n characters from a string if you maybe want to include it, It does something but a different way around.

_makhmud says

January 17, 2022 at 3:46 pm

14-exercise

n = int(input("Enter number:"))

while n > 0:
    for i in range(1, n+1):
        print("*", end = "")
    print("\t")
    n -= 1

kandula nithin says

February 8, 2022 at 7:13 pm

num = int(input("please enter the number of lines needed: "))
while num > 0:
  print('*' * num, end = '')
  num -= 1
  print(end = '\n')

Arnav(9 years old) says

January 14, 2022 at 10:08 pm

def exponent(b,e):
  b = b
  e = e
  a = 1


  for i in range(e):
    a = a*b

  print(b,'to the power of',e,'is',a)
x = int(input('enter a base to caculate exponents: '))
y = int(input('enter exponent(how many times you multiply): '))
exponent(x,y)
print(' ')
print(' ')
for i in range(5,0,-1):
  print('* '*i)

Arbaz Khan says

January 12, 2022 at 5:24 pm

Last Question We Can Also Do Like:

def num(x, y):
  return x**y

num_1 = int(input("ENTER THE BASE: "))
num_2 = int(input("ENTER THE EXPONENT: "))
RESULT = num(num_1, num_2)
print("Result: ",num_1,"raises the power of ",num_2,"=",RESULT)

Arnav(9 years old) says

January 12, 2022 at 8:53 am

def exponent(b,e):
  b = b
  e = e
  a = 1


  for i in range(e):
    a = a*b

  print(b,'to the power of',e,'is',a)
x = int(input('enter base(for exponents): '))
y = int(input('enter exponent(how many times you multiply): '))
exponent(x,y)

#Exercise 15

Arnav(9 years old) says

January 12, 2022 at 8:55 am

indents didn’t post when i copy pasted my code

Reply

Arnav(9 years old) says

January 4, 2022 at 12:02 pm

var = input('enter a word: ')
#print(var)
x = int(input('enter a number that is less than len of the word you typed: '))
while x<=len(var)-1:
  print(var[x],end='')
  x += 1
for i in range(2,5):
  print(' ')

Arnav(9 years old) says

January 4, 2022 at 12:06 pm

this is for exercise 4, the one about removing index chars

Reply

David says

December 21, 2021 at 1:56 am
Please, someone, tell me why my solution for #12 doesn’t give me the correct result for 45000 income because it will always tell me that the tax is 3500 instead of 6000:
```
def tax(income):
    if income  10000  20000:
        print("Tax to be payed: ", (income-20000)*0.2+1000)

tax(5000)
tax(15000)
tax(45000)
```
Reply
- David says
  
  December 21, 2021 at 1:57 am
  sorry wrong code:
```
def tax(income):
    if income  10000  20000:
        print("Tax to be payed: ", (income-20000)*0.2+1000)
```
  Reply
  - David says
    
    December 21, 2021 at 1:59 am
    
    Ok for some reason it won’t copy/paste the whole code correctly… FFS
    
    Reply
    - David says
      
      December 21, 2021 at 2:02 am
      
      Last try, very sorry for spamming:
      
      def tax(income): if income 10000 20000: print("Tax to be payed: ", (income-20000)*0.2+1000) tax(5000) tax(15000) tax(45000)
      
      Reply
      - David says
        
        December 21, 2021 at 2:08 am
        
        Nevermind I figured it out. My comments can be deleted. Again very sorry for spamming, this was not my intention.
    - Vishal says
      
      January 2, 2022 at 11:12 am
      
      Please use ‘pre’ tag for posting code. For example,
      
      Your code here
      
      Reply
- Irfan says
  
  May 15, 2023 at 12:14 pm
  
  first = 10000 second = 10000 remaining = tax - (first + second)
  x = (first * 0/100 + second * 10/100 + remaining * 20/100) print("The taxable amount is : ", x)
  
  Reply

hakizimana frederick says

December 15, 2021 at 5:19 pm


def merge_list(list1, list2):
    empty_arr = []

    for el1, el2 in zip(list1, list2):
        if el1 % 2 == 0:
            empty_arr.append(el1)
        if el2 % 2 != 0:
            empty_arr.append(el2)
    return empty_arr


print(merge_list(list1, list2))

hakizimana frederick says

December 12, 2021 at 1:53 am


input_str = str(input('Enter string: '))

for i,  value, in enumerate(input_str):
    if i % 2 == 0:
        print(value)

Jakub Kolář says

December 11, 2021 at 12:04 am

Hello. The query for task 13. Why is print("\t\t") at the end? When does print("") work anyway?

Reply
- Rahul says
  
  January 16, 2022 at 11:35 pm
  
  print("\t") is used whenever a tab space has to be left in the output
  print("") is used to print a new line
  
  Reply

Saeed Ahmed says

November 25, 2021 at 11:04 pm

original_str = input("Enter sTring : ? ")
remove_chr = int(input("Enter number of chr to remove :? "))
new_str = original_str[remove_chr:]
print(new_str)

# Method 2

original_str = input(" Enter sTring : ? ")
remove_chr = int(input("Enter number of chr to remove :? "))
new_str = list(original_str[remove_chr:])
print(new_str)

# Method 3
original_str = input(" Enter sTring : ? ")
remove_chr = int(input("Enter number of chr to remove :? "))
new_str = list(original_str)
new_str = new_str[remove_chr:]
print(new_str)

Aakash Thakur says

November 24, 2021 at 2:36 pm

for question number 15.

base = 2
exponent = 5
for i in range(1,exponent+1):
    ans = base**exponent
    print("{} raises to the power of {}: {}".format(base,exponent,ans))
    break

Ankit Mondal says

December 14, 2021 at 7:36 am

def power(base, exponent):
    exp = (base**exponent)
    print(f'{base} raises to the power of {exponent} is {exp}')


new_power = power(3, 2)

Ankit Mondal says

December 14, 2021 at 7:37 am


def power(base, exponent):
    exp = (base**exponent)
    print(f'{base} raises to the power of {exponent} is {exp}')

new_power = power(3, 2)

Test says

January 6, 2022 at 8:50 pm

Test

Reply

Michael says

November 16, 2021 at 12:35 am

Q3: An overly complex, but fun, answer

string = 'abcdefghijklmnopqrstuvwxyz'
def slicer(x):
    for i in range(len(x)):
        if i % 2 == 0:
            print(string[i])

print(slicer(string))

Coder says

November 1, 2021 at 5:31 pm

#Exercise 14: Print downward Half-Pyramid Pattern with Star (asterisk)
for i in range(1,6):
    print("* "*(6-i))

Mohammed Sadique Ansari says

October 12, 2021 at 7:49 pm

For Ex:15

Reply
- Mohammed Sadique Ansari says
  
  October 12, 2021 at 7:54 pm
```
def exponent(base, exp):
                output = base**exp
                return output
```
  Reply
Jim says

September 7, 2021 at 1:39 am
EX 1:
“Given two integer numbers return their product only if the product is greater than 1000, else return their sum.”

Hint:
Next, use the if condition to check if the product >1000. If yes, return the product
Otherwise, use the else block to calculate the sum of two numbers and return it.

Your solution:
```
if product <= 1000:
        return product
    else:
        # product is greater than 1000 calculate sum
        return num1 + num2
```
Your solution is the opposite of what your problem statement and your hint state.
Reply
- mustafa says
  
  September 12, 2021 at 8:48 am
  
  yes nice catch
  
  Reply
- John says
  
  November 4, 2021 at 12:11 pm
  
  Came here to say the same. Already unconfident beginners get even more confused with this mistake in the very first exercise.
  
  Reply
  - Quan says
    
    November 10, 2021 at 1:45 pm
    
    absolutely agree with you
    
    Reply
  - Jean says
    
    November 12, 2021 at 12:22 am
    
    Error still not corrected as of November 11. This is really a poor site, they don’t even read comments… Shame on them !
    
    Reply
- JJ says
  
  January 1, 2022 at 2:56 pm
  
  Right, it is a bit amusing that is the first exercise. Just a side note: shouldn’t it be <, not <= because the question was: "if the product is greater"?
  
  Reply

Houssem says

September 6, 2021 at 5:31 pm

Q 12:

def stars(num):
    for i in range(num+1, 0, -1):
        print("* "*i)
    
stars(5)

Houssem says

September 6, 2021 at 4:15 pm

Q 11:

def number(num):
    string = str(num)[::-1]
    for i in string:
        print(i, end=" ")

Houssem says

September 6, 2021 at 4:05 pm

Q 10:

def makeList(lst1, lst2):
    f_lst = list()
    for i,a in zip(lst1, lst2):
        if i%2 != 0: f_lst.append(i)
        if a%2 == 0: f_lst.append(a)
    return f_lst

Houssem says

September 5, 2021 at 10:17 pm

Q 09 in two lines:

def reverse(num):
    a = int(str(num)[::-1])
    return True if a == num else False

Houssem says

September 5, 2021 at 9:52 pm

Q7 without using str class:

def emma(string):
    lst = string.split(" ")
    count = 0
    for i in lst:
        if i == "Emma": count += 1
    retutn count

kanhaiya singh says

August 24, 2021 at 6:44 pm

solution for question no 5

if list[0]==list[-1]:
    print('true')
else:
    print('false')

Houssem says

September 5, 2021 at 9:22 pm

This is more easy I think:

def q5(lst):
     return True if lst[0] == lst[-1] else False

badr says

September 21, 2021 at 2:31 pm

num1 = [10, 20, 30, 40, 10]
num2 = [75, 65, 35, 75, 30]


def detective(nums):
    if (nums[0]) == (nums[-1]):
        return True
    else : 
        return False
    
print(detective(num1))
print(detective(num2))

Meiti says

August 20, 2021 at 7:06 am
other way to coding ex 8:
```
for i in range(num):
    print((str(i) + ' ') *i) 
```
Reply

Meiti says

August 20, 2021 at 6:26 am

Other way to solution Q5:

def same(x):
    y=[]
    y.append(x.pop(0))
    y.append(x.pop(-1))
    k=set(y)
    if len(k)==1:
        return True
    else:
        return False

a little bit complicated

Aditya Sarkar says

August 27, 2021 at 12:06 am

Another way of Question 5:

def alist(list):
    for i in list:
        if i[0] == i[-1]:
            return True
        else:
            return False

print(alist([[10, 20, 30, 40, 10]]))

Aditya Sarkar says

August 27, 2021 at 12:08 am

probably not the best looking code

Reply
- Houssem says
  
  September 5, 2021 at 9:24 pm
  what about this one?
```
def muli(lst):
     return True if lst[0] == lst[-1] else False
```
  Reply

Meiti says

August 20, 2021 at 4:28 am

Another answer to E3:

def evenchr(str):
	x=list(str)
	for i in x[0::2]:
		print(i)

Collince says

November 1, 2021 at 1:34 am

def evenChar(text):
	for i in text[0::2]:
		print(i)

Viraj Sazzala says

August 19, 2021 at 10:29 am

Question no: 11 – Reverse a number
this is an easier and less complicated way to do it.

number = input('Enter a number: ')

#reverse the string input
reverse_num = number[::-1]

#turns the reverse number into a list
list_num = list(reverse_num)

#print the list with spaces in the same line
print(*list_num)

Hoang Thien Son says

August 18, 2021 at 3:42 pm

dear admin this is my shorter solution to question 8

n = int(input("Your number of rows: "))         #call the rows
for i in range(n+1):    #for i in 1 to n
    print(str(i)*i)     #print the number of i multiple by i which is the same point of this exercise

Viraj Sazzala says

August 18, 2021 at 12:04 am
Question 9: Reverse number.
```
number = input('Enter a number: ')
revnum = number[::-1]                            
if number == revnum:
    print('True')
else:
    print("False")
```
this is simpler and also works well.
Reply
- Lynnz says
  
  September 19, 2021 at 2:07 pm
  
  The same with you!
  
  Reply
- badr says
  
  September 21, 2021 at 3:03 pm
```
x = list(input("your number :"))
if x[0::] == x[-1::-1] :
    print("Yes. given number is palindrome number")
else : 
    print("No. given number is not palindrome number")
```
  Reply
  - Andy says
    
    October 14, 2021 at 7:36 pm
    Assuming n is a number:
    
    def is_palindrome(n): return (list(str(n1)) == list(str(n1))[::-1])
    
    It’s, as mentioned before, a one-liner.
    Badr, [:] is enough to copy ([0::] is not needed, and moreover, you don’t need to copy the list). and x[::-1] is enough to return a reversed copy. Just to polish a bit the already efficient code 🙂
    Reply
    - Andy says
      
      October 14, 2021 at 7:37 pm
      
      Damn… forgot to update n1 to n in the body of the method…
      
      Reply
- Andy says
  
  October 14, 2021 at 7:39 pm
  
  Hey, missed this one. Nice.
  
  Reply

Adam says

August 17, 2021 at 1:24 pm

Hi Admin,
thanks, I’m still very new to programming and these exercises improve my skills 🙂

my answer for Q3 & Q10:

def print_even_index(input_str):
    char_at_even_index = [(i, j) for i, j in enumerate(input_str) if i%2 == 0]

    for i, j in char_at_even_index:
        print("index[", i, "]", j)

input_str = "pynativeZ"

print_even_index(input_str)

————-

def merge_list(list1, list2):
    list1_odd_nums = [i for i in list1 if i%2 == 1]
    list2_even_nums = [i for i in list2 if i%2 == 0]
    new_list = list1_odd_nums + list2_even_nums
    return new_list

list1 =  [10, 20, 23, 11, 17]
list2 = [13, 43, 24, 36, 12]

print(merge_list(list1, list2))

Voided says

August 5, 2021 at 12:59 am

My solution to the very easy first problem but with less lines of code:


# problem 1:
def calculate(n1,n2):
    if n1 * n2 >= 1000:
        return n1 + n2
    else:
        return n1 * n2

result = print(calculate(int(input("1st number: ")),int(input("2 number: "))))

Madhan hasee says

July 31, 2021 at 2:23 pm

You can use this code for 15 Exercise,

def exponent(base, exp):
     return base ** exp
print("6 raises to the power of 4:",exponent(4, 6))

Madhan hasee says

July 31, 2021 at 2:25 pm

def exponent(base, exp):
       return base ** exp
print(“6 raises to the power of 4:”,exponent(4, 6))

angelo says

August 4, 2021 at 6:26 am

def exponent(base,exp):
    if base > 0:
        print(base**exp)
exponent(2,5)

Mazhar Khilji says

August 4, 2021 at 2:23 pm

def exponent(base, exp):

    power = base ** exp
    print(base, "raises to the power of", exp, "is: ", power)

exponent(5, 4)

Madhan hasee says

July 31, 2021 at 2:17 pm

For 11 Exercise you can use this code,

numbers = input("Enter your number :")
for i in numbers[::-1]:
     print(i, end=' ')

Madhan hasee says

July 31, 2021 at 2:18 pm

numbers = input(“Enter your number :”)
for i in numbers[::-1]:
       print(i, end=’ ‘)

Mazhar says

July 17, 2021 at 3:28 pm

Exercise 15. will be easy like this

def powerplay(base,exponent): 

    while True:
        return base ** exponent

print("2 raises to the power of 5:",powerplay(2,5))

ANKIT says

July 26, 2021 at 4:28 pm

def result(x,y):
    return x**y
print(result(2,5))

Ritvik says

June 12, 2021 at 2:55 am

Another way to solve Q14 ->

for i in range(5, 0, -1):
    print(' * '*i, end='')
    print('\t')

Ritvik says

June 12, 2021 at 2:56 am


for i in range(5, 0, -1):
    print(' * '*i, end='')
    print('\t')

Piyush says

July 2, 2021 at 3:52 pm

i=6
while i>0:
  print("* " * i)
  i -= 1

Shreyans Daga says

July 14, 2021 at 7:58 am

Another way could be:

count = 4
string = "* * * * *"
list1 = string.split(" ")
for i in range(1, 6):
    print("".join(list1))
    list1.pop(count)
    count -= 1

Stevie says

June 6, 2021 at 12:47 am
Exercise 8:
The below will also return the same result.
```
for i in range(6):
    print(str(i)*i)
```
Reply
- t says
  
  June 18, 2021 at 3:56 am
  
  pattern is wrong
  
  Reply
  - Gaurav Gautam says
    
    July 19, 2021 at 6:00 pm
    
    just having a 0
    it should be range(1,6)
    rest all is perfect
    
    Reply

Reshma Rajan k says

June 4, 2021 at 9:38 am

Exercise :15

def exponent(base,exp):
    print('Base=',base,'exponent=',exp)
    ans = base**exp
    print(exp,'raises to the power of',base,':',ans)
exponent(5,4)

Subhraneel Paul says

May 23, 2021 at 7:32 pm
Exercise 14: Print downward Half-Pyramid Pattern with Star (asterisk)
```
for i in range(5):
    for j in range(5-i):
        print("*", end="  ")
    print("\n")
```
Reply
- Subhraneel Paul says
  
  May 23, 2021 at 8:00 pm
```
for x in range(5):
    for y in range(5-x):
        print("*",end="  ")
    print("\n")
```
  Reply
- AndyLee says
  
  June 6, 2021 at 7:41 am
  
  Thank you! This one is easier for me to understand!
  
  Reply
  - Subin Jerin says
    
    July 15, 2021 at 6:01 pm
    
    Very easy to understanf for me this
    
    Reply

Gulshat says

May 14, 2021 at 12:22 pm

Ex9: I prefer to treat entered number as a string. Way more simple code:

s = input("Input a number: ")
if s[::-1] == s:
    print("The original and reverse number is the same")
else: 
    print("The original and reverse number is not the same")

Gulshat says

May 14, 2021 at 12:12 pm
Hi, I have a slightly shorter solution for Ex8:
```
for i in range(1, 10):
    print (str(i)*i)
```
Reply
Tawhid says

May 13, 2021 at 3:03 pm

Hello vishal
could you elabaorate this line of code for me ?
reverseNum = (reverseNum * 10) + reminder

It was in Q9. I am new in python so some easy things seems hard for me.The line is one of them.

Reply

Abhay MISHRA says

May 9, 2021 at 11:13 pm

Ans-1

def pruduct_sum(num1,num2):
  if num1*num2 > 1000:
    return num1+num2
  return f'The result is {num1*num2}'
print(pruduct_sum(40,30))

Ans-2:

for i in range(10):
  if i == 0:
    print(f'current number is {i} no previse number : sum{i}')
  else:
    print(f'current number is {i} previsios number is {i-1} : sum {i+(i-1)}')

Ans_3:

def reverse_num(intiger):
  new_num = str(intiger)
  if new_num == new_num[::-1]:
    print(f'Orignal number is {intiger}','\n''The original and reverse number is the same')
  else:
    print("The original and reverse number is not same")

reverse_num(121)

Ans-4:

str = "pynative"
for word in str[::2]:
  print( f'{str.index(word)} : {word}')

Ans-5

def removeChars(var,num):
  return var[num:]
print(removeChars("pynative", 4))

Ans-6

def Emm_counter(string):
  total = 0
  new_list = string.split(' ')
  for item in new_list:
    if item == "Emma":
      total += 1
  return total
print(f'Emma appeared  {Emm_counter("Emma is good developer. Emma is a writer")} time')

Ans-7

for i in range(1,6):
  for num in range(i):
    print(i , end=' ')
  print()

Ans-8

def even_odd_list(list1,list2):
  new_list = []
  for num in list1:
    if num%2!=0:
      new_list.append(num)
  for num in list2:
    if num%2 == 0:
      new_list.append(num)
  return new_list
list1 =  [10, 20, 23, 11, 17]
list2 = [13, 43, 24, 36, 12]
print('new_list is ', even_odd_list(list1,list2))

Ans-9

def reverse_num(intiger):
  new_string = ''
  new_num = str(intiger)
  for num in new_num:
    new_string += num+' '
  return new_string[::-1]

print(reverse_num(7536), end=" ")

Ans-10

def income_tax(income):
  tax_payble = 0
  if income = 10000:
    tax_payble += 10000*10/100
    first_year -= 10000
    tax_payble += first_year*20/100
  else:
    if  first_year < 10000:
      tax_payble += first_year*10/100
  return tax_payble
print(income_tax(100000))

Ans-11

for i in range(1, 11):
    for j in range(1, 11):
        print(i * j,end=' ')
    print('\n')

Ans-12

for i in range(1,6):
  print('* '* (6-i))

Ans-13

def exponent(base, exp):
  var = base**exp
  str_bas = '('+str(base)+ (' *'+str(base))*(exp-1)+' = '+str(var)+')'
  power_vale = (f'{base} to power of {exp } : {var}')
  print(power_vale,'\n',str_bas)

exponent(2,5)

Sab3rson says

February 12, 2022 at 8:45 am

I got a similar answer:

def exponent(base,exp):
    
    product = base ** exp
        
    answer_str = "("+ str(base) + (" * " + str(base)) * (exp -1) + " = " + str(product) +")"
        
    print(f"base = {base}\nexponent = {exp}\n\n{base} raises to the power of {exp}: {answer_str}")
    
exponent(2, 5)

Coconut says

May 8, 2021 at 6:23 am

Ex 15 alt:

def exponent(base,exp):
	## Gives base to the power of exp using only multiplication
	result = 1

	print(base, 'to the power of', exp, 'is: ')

	for i in range(exp):
		result = result*base
		if i<exp-1:
			print(base,'x', end=' ')
		else:
			print(base, end=' ')

	print('=',result,end =' ')



exponent(2,3)

Piyush says

July 2, 2021 at 4:00 pm

this is the sort version

def exponent(base, exp):
  return(base**exp)

print(exponent(5,4))

Coconut says

May 8, 2021 at 6:21 am

EX 15: alt

def exponent(base,exp):
	## Gives base to the power of exp using only multiplication
	result = 1

	print(base, 'to the power of', exp, 'is: ')

	for i in range(exp):
		result = result*base
		if i<exp-1:
			print(base,'x', end=' ')
		else:
			print(base, end=' ')

	print('=',result,end =' ')

exponent(2,8)

Dave Perez says

May 4, 2021 at 7:34 pm

Thank you for all these exercises. They were very helpful to practice coding in Python.

Here are alternative solutions for Exercise 15:

def exponent(base, exp):
    # built-in by python exponent function
    base = base**exp
    return base
print(exponent(2, 5))

def exponent(base, exp):
    # own logic
    old_base = base
    for index in range(exp-1):
        base = base * old_base
    return base
print(exponent(2, 5))

Dave Perez says

May 4, 2021 at 6:57 pm

Alternative and shorter solution in Exercise 12.

def exercise12(salary):
    tax = 0  # default tax is 0 for salary 10K and below
    if salary > 20000:  # salary more than 20K is 1000 + 20% of salary exceeding 20K
        tax = (salary-20000) * 0.2 + 1000  # 1000 is 10% of 10000
    elif salary > 10000:  # salary more than 10K is 10% of salary exceeding 10K
        tax = (salary-10000) * 0.1
    return tax

# Remove comment to run exercise12
#print(exercise12(45000))

Dave Perez says

May 4, 2021 at 6:39 pm

Alternative solution for Exercise 11.

def exercise11(num):
    reverse = -1
    for index in range(len(str(num))):  # loop in range of string length
        print(str(num)[reverse], end=" ")  # print using negative index
        reverse -= 1  # decrement to next negative index
# Remove comment below to run exercise11
#exercise11(7536)

Dave Perez says

May 3, 2021 at 2:18 pm
Found a shorter solution for Exercise 9:
Exercise 9: Reverse a given number and return true if it is the same as the original number
```
def exercise9(num):
    if str(num) != str(num)[::-1]:
        return False
    return True
# Remove comment below to run exercise9
#print(exercise9(121))
```
Reply
Dave Perez says

May 3, 2021 at 1:26 pm
First of all, thank you for these exercises.

The solution in Exercise 3 does not print the last character despite having an even index. Improved solution provided below:

Exercise 3: Given a string, display only those characters which are present at an even index number.
```
def exercise3(word):
    for index in range(0, len(word), 2):
        print(word[index])

# Remove comment below to run exercise3
# exercise3("pynative")
```
Reply
pramod padhye says

April 25, 2021 at 11:36 am
can we use below code for exc.8
```
for i in range(6):
    i = str(i)*i
    print(i)
```
Reply
african bum disese says

April 23, 2021 at 1:51 pm

same i made roblox epic games and gta 5 im making gta 6 in pythons

Reply
subramani says

April 22, 2021 at 9:33 am

Hi Vishal,

First of all, thank you for the exercises, it is easy to understand and refer relevant docs when needed.

I have a question regarding exercise no 4, how do I print first 4 char only, with above code, it only displays in reverse order. My expected output would be pyna, how do I achieve it

Reply

Rishu says

April 14, 2021 at 9:15 pm

Solution 15 –


a=int(input("Enter base : "))
b=int(input("Enter exponent : "))

def exponent(base,exp):
    return (f"{base} raises to the power of {exp} : {base**exp}")

print(exponent(a,b))

Rafael says

April 14, 2021 at 4:53 am

Thank you!
You may like this for the 15th exercise. I went a bit further.


def exponent(base, exp):
    p = base**exp
    print(str(base) + " raises to the power of " + str(exp) + ": " + str(p) + " i.e. (", end = " ")
    for i in range(exp):
        print(base, end= " ")
        if i < exp - 1:
            print("*", end = " ")
    print("= " + str(p) + ")")
exponent(5, 4)

jeevan kumar says

March 31, 2021 at 7:05 pm

Good work sir! basic questions but for a beginner, it is a good start. I have done all 15 questions and felt good. thank you. I am from CBIT, if you find any job for me, please let me know. I am a software developer.

Reply
Jasper Chima says

March 22, 2021 at 5:52 pm

Although i used the "".join(reversed('my_num" ) #if you noticed, my_num was a str and not an int so i need to understand your method.

Reply
jasper chima says

March 22, 2021 at 5:47 pm
Please, I do not understand the logic of your solution to Exercise number 9: Reverse a number and return True or False. Below is your code and by the side is how I understand it.
I need assistance as I’m very new to programming.
```
def reverseCheck(number):
    print("original number", number)  # let number = 121
    originalNum = number  # original number = 121.
    reverseNum = 0  # reverseNum  = 0
    while (number > 0):  # while 121 > 0:
        reminder = number % 10  # remainder = 121% 10, therefore remainder == 1.
        reverseNum = (reverseNum * 10) + reminder  # reverseNum = (0 * 10) + 1, therefore reverseNum == 1
        number = number // 10  # number = 121 // 10, therefore number == 12
    if (originalNum == reverseNum):  # if (121 == 1):
        return True  # return True . How does 121 equals 1? i'm confused
    else:
        return False


print("The original and reverse number is the same:", reverseCheck(121))
```
Thanks for your clarification in advance. I love your webpage. God bless and increase you.
Reply
- Jasper Chima says
  
  March 22, 2021 at 7:23 pm
  
  Dear founder o Pynative,
  Sorry for disturbing this comment section with questions.
  As regards my earlier question to exercise 9, I now fully understood the logic behind your solution.
  Exercise 11 helped me better. The comments (#) you included were very pivotal to my understanding.
  Thank you once again.
  
  Reply
Cholavendhan says

March 19, 2021 at 2:32 pm

Wonderful set of examples.. Great Work dude…

Reply
- Vishal says
  
  March 21, 2021 at 11:24 am
  
  Thank you, Cholavendhan.
  
  Reply
Avinash Reddy says

March 16, 2021 at 1:51 am

Must Appreciated .
Nice excercises

Reply
- Vishal says
  
  March 16, 2021 at 11:36 am
  
  Thank you, Avinash.
  
  Reply
**Andre : ) : ) says

March 14, 2021 at 11:28 pm

Thank u sir for the exercises….. : )

* Andre*

Reply
Khan says

March 14, 2021 at 11:23 pm

Really helpful for PYTHON beginners… pls add more resources like this

Reply

Rashin says

March 14, 2021 at 11:14 pm

Exercise 14


n = 5
while n>=1 :
    print('* '*n,'\n')
    n -= 1

Rashin says

March 14, 2021 at 11:08 pm
#EX 4
```
txt = 'pynative'
n = 4
print( txt [ n : ] )
```
Reply
Rashin says

March 14, 2021 at 11:07 pm

Thank u sir

Reply
Satheesh says

March 13, 2021 at 8:33 am
Q3:
```
print(given_string[1::2])
```
Reply
Petr says

March 11, 2021 at 6:57 pm

Exercise 3
If you have text with 7 letters (length=7) the index of last letter is [6].
So in order to get it printed you need range from 0 to 7 (length) as the 7 is not included.
Having the length – 1 as in the solution is not correct.

If you would remove the “e” from Pynative – the “v” with index [6] wouldn’t get printed as the range would be from 0 to 6 and 6 wouldn’t be included.

Just minor detail 🙂

Reply
simanta says

February 23, 2021 at 8:40 pm

great sir. really helpful for beginners

Reply
- Vishal says
  
  March 2, 2021 at 8:54 pm
  
  You’re Welcome.
  
  Reply
Matt says

February 17, 2021 at 12:41 am

There is something wrong with your commenting system. I try to hit “reply” and it does nothing. Could it be a browser plugin or something?

Reply
- Vishal says
  
  February 19, 2021 at 7:55 am
  
  Hey Matt, It is working as expected. Can you please try again.
  
  Reply
Matt says

February 16, 2021 at 2:30 am

Greetings,

I have just started learning Python. However, I have experience in other OOL. I have a few issues with your examples.

1. It’s difficult what you’re expecting to see as output. For instance, #12. The output you have there has two spaces between the initial integer and it’s product. So, I was trying to format that as such and it was starting to get messy. Your solution is what I would have done if I didn’t’ expect to have two spaces.

The same applies to the tax rate problem. I thought that is what you wanted to see for output. But, obviously you just wanted the number.

The examples with the reverse comparisons where easy to do with slicing, but you had this over complex way of taking an integer and using multiple loops to reverse the value. I just converted them to strings and used slicing. Then, back to integers.

It’s been very helpful and I will continue to use it. However, I don’t think you should be presumptuous and expect someone to know exactly it is you’re looking for.

Just my 2 cents.

Reply
- Iván says
  
  January 1, 2023 at 2:42 am
  
  agree
  
  Reply
Vanishri says

February 5, 2021 at 11:39 am

2 and 3 ques “int object is not callable”

Error is displaying

Reply

Gurpreet singh says

January 30, 2021 at 2:09 pm


li=[1, 45,3,55,1]
if li[0]==li[4]:
     print("True") 
else:
        print("False")

Iwanttolearn says

January 28, 2021 at 1:56 pm

Hi I’m also a Beginner, This is my second day of reviewing some Exercises.
I can’t even type any single line code.

Reply
MJ says

January 25, 2021 at 12:23 pm

Please type the code for the following:-

1)
Input three city names from the user. Store these names in a list. Shift all the three names to a single variable and delete the list.
2)
Input three numbers from the user and create a list containing these numbers. Insert number 25 at the second position and finally calculate the product of all the elements of the list.

Reply

Prayas says

January 21, 2021 at 8:20 pm

Question no.15 has a better answer XD


def exponenet (base, exp):
    x=base**exp
    print(str(base), "raised to the power", str(exp), "is : ", x)
exponenet(6,2)

Prayas says

January 21, 2021 at 8:18 pm

Question no 15 has a better solution XD:

def exponent (base, exp):
    x=base**exp
    print(str(base), "raised to the power", str(exp), "is : ", x)
exponent(6,2)

yahia mahmoud says

January 12, 2021 at 12:39 am
The most helpful site ever learned a lot, thanks.
———————–
```
The_most_helpful_site_ever = int(len(input("insert every word in pynative.com ")))
for i in range ( The_most_helpful_site_ever) :
    print("thanks ", end = "")
```
Reply
- Vishal says
  
  January 12, 2021 at 8:37 am
  
  Thank you, Yahia. I am glad you liked it.
  
  Reply

Amrita Mehta says

December 27, 2020 at 5:01 pm

Question 5


l = [10, 20, 30, 40, 10]
if l[0] == l[-1]:
    print("True")
else:
    print('False')

Akang_senju says

August 5, 2021 at 6:13 pm

l = [10, 20, 30, 40, 10]
print( True if l[0] == l.pop() else False)

Amrita Mehta says

December 27, 2020 at 4:45 pm

Question 1

def calc(n1,n2):
    product = n1 * n2
    
    if product > 1000:
        return n1 + n2
    else:
        return product
    
print(calc(3000,2))

Nidhi Mehta says

December 27, 2020 at 4:56 pm
Q3.
```
str1 = input('enter string:')
str1[::2]
```
Reply

Rajkumar says

December 17, 2020 at 8:47 pm
Question 14: Print downward Half-Pyramid Pattern with Star (asterisk)
```
for i in range(5, 0,-1):
    print(' *' * i)
```
Reply
Noname says

November 11, 2020 at 6:55 pm
Hi! I am a beginner in Python, so thanks a lot for the exercises. I have got a question concerning exercise 3. I have written the following code:
```
 
def thatsweird(a):
    for x in a :
        if a.index(x)%2==0:
            print (x)
            
a= list(input("word: " ))
thatsweird(a)
```
It works fine as long as there are no identical letters in a row in which case it messes up. I would be very grateful if someone explained to me why or just referred me to an explanation online as I didn’t find one (yet):)

Thanks!
Reply
- Z09 says
  
  November 14, 2020 at 7:50 pm
```
string = input("Enter your word: ")
print("Original string is " + string)
print("Printing only even index chars")
for i in range(len(string)):
    even = i%2
    if even == 0:
        print("index["+ str(i) + "] " + string[i])
print("NOTE:THAT SPACE ALSO COUNTS AS A CHAR\n\tTHE COUNT STARTS FROM ZERO NOT ONE")
```
  Reply
- Priyank Palshetkar says
  
  November 24, 2020 at 4:57 pm
  Easy solution for exercise 3
```
str="pynative"
for i in range(0,len(str)-1,2):
    print(str[i])
```
  Reply
- ayush says
  
  December 14, 2020 at 10:10 am
```
name='umbrella'
for i in range(0,len(name)-1,2): 
    print(name[i])
```
  Reply
- Multicast says
  
  January 10, 2021 at 3:21 am
  Hi Noname, Whenever you try to find an index for a character Python would return the index where ever it finds it first, so if your string has an identical char Python will just give you the index for the first occurring char. hope it helps.
  
  Below is how I approached this problem:
```
def even_str(my_str):
    n = len(my_str)
    for i in range(0, n, 2):
        print(my_str[i])
even_str("pynative")
```
  Reply
suraj says

November 11, 2020 at 4:47 pm
# # Question 14: Print downward Half-Pyramid Pattern with Star (asterisk)
```
n=5
for i in range(6):
#     print(i)
    z= "* " * n
    n= n-1
    print(z)
```
Reply

Rushikesh says

November 2, 2020 at 3:07 pm

Que 5:

def isFirst_And_Last_Same(numberList):
    print("Given list is ", numberList)
    firstElement = numberList[0]
    lastElement = numberList[-1]
    if (firstElement == lastElement):
        return True
    else:
        return False

My_list = []
n = int(input('number of elements = '))
for i in range(0,n):
    ele = int(input(f'ele {i+1} = '))
    My_list.append(ele)

Sebz says

November 27, 2020 at 4:59 am

Easy solving for Q. 5

list1 = [10, 20, 30, 40, 10]

bool = list1[0] == list1[-1]

print(f'The result is {bool}')

Rob says

October 28, 2020 at 7:50 pm

Question 13:

Why is it print("\t\t") and not print("\n")?
I thought \t creates a tab or indent. \n drops a line doesn’t it?
Alos why \t\t and not just \t

Thanks. I enjoyed the questions finding some pretty tricky!!!

Reply
- Lokesh says
  
  November 9, 2020 at 1:52 pm
  
  \t\t means is two tab
  
  Reply

Attiya says

October 23, 2020 at 5:14 am

def asterisks(input):
     if input > 0:
         for y in range(input,0,-1):
              print('*'*y)

jamar says

October 21, 2020 at 8:28 pm

rows=5
for x in range(rows,0,-1):
    print('*'*x)
    print('\n')

suraj says

November 11, 2020 at 4:45 pm

I have tried this way

n=5
for i in range(6):
#     print(i)
    z= "* " * n
    n= n-1
    print(z)

Sindhu says

October 21, 2020 at 7:53 pm

For question 6,we can have like this,


def CompareLists(givenList):
    return (givenList[0]== givenList[len(givenList)-1]) 
list1 = [10, 20, 30, 40, 10]
list2 = [10, 20, 30, 40, 50]
print(CompareLists(list1))
print(CompareLists(list2))

Supriya says

October 19, 2020 at 7:56 pm
Question 15 can have a simpler solution . Which is :
```
def exponent(base, exp):
    return f'{base} raises to the power of {exp} is {base**exp}'

print(exponent(5, 4))
```
Reply
- Sindhu says
  
  October 23, 2020 at 7:54 pm
  
  Yeah!!.. That’s what am going to say!!
  
  Reply
Abdu says

October 16, 2020 at 10:26 pm
question 14
```
 for x in range(6,0,-1):
    print(x*" *")
```
Reply

Abdu says

October 16, 2020 at 10:23 pm

question 15

def exponent (base,exp):

    return base**exp


print(exponent(5,4))

Abdu says

October 16, 2020 at 9:26 am

question 11.

num = int(input("Enter any number: "))
reverse = 0
lst = []
while num > 0:
    reminder = num%10
    reverse = (reverse*10)+reminder
    num = num//10
    lst.append(reminder)


print(*lst,sep=" ")

Abdu says

October 16, 2020 at 9:28 am

 num = int(input("Enter any number: "))
reverse = 0
lst = []
while num > 0:
    reminder = num%10
    reverse = (reverse*10)+reminder
    num = num//10
    lst.append(reminder)


print(*lst,sep=" ")

corleone1986 says

June 27, 2022 at 11:20 pm

Use ‘pre’ tag for posting code. E.g.

 def checking(num):
    x = str(num)[::-1].replace("" , " ")
    print(x)

checking(7658)

Attiya says

October 10, 2020 at 5:05 am
Simplified solution for question 14 without nested loop:
```
rows=5
for x in range(rows,0,-1):
    print('*'*x)
    print('\n')
```
Reply

Isaac says

October 9, 2020 at 1:35 am

Question 3


cv=0

str="pynative"
for s in str:
    if cv %2==0:
        print(s)
    cv+=1

Isaac says

October 9, 2020 at 1:31 am

Question 1


    num1=20
    num2=30
    product=num1*num2
    if product > 1000:
        print(num1 + num2)
    else:
        print(product)

Isaac says

October 9, 2020 at 1:25 am

Question 12


    def exponent(base,exp):
        return base**exp
a=exponent(2,5)
print(a)

Emerson says

September 22, 2020 at 12:14 pm
The question to exercise 6:
Why does a “None” appear at the end of my output?

Here is my code:
```
def onlyDivisible(numbersList):
    print("given list is", numbersList)
    print("Divisible of 5 in a list")
    for n in numbersList:
        if n % 5 == 0:
            print(n)


myList = [10, 20, 33, 46, 55]
print(onlyDivisible(myList))
```
Reply
- Sumanth says
  
  September 24, 2020 at 10:38 am
  
  Hello Emerson
  Instead of using use
  
  Reply
- Sumanth says
  
  September 24, 2020 at 10:41 am
  
  instead of using use
  
  Reply
- Sumanth says
  
  September 24, 2020 at 10:43 am
  
  Instead of using print(onlyDivisible(myList)) use onlyDivisible(myList)
  
  Reply

Sumit says

September 11, 2020 at 1:11 pm

Question 15: Write a function called exponent(base, exp) that returns an int value of base raises to the power of exp.

def exponential(base,exp):
    return(f"{base} raise to power {exp}: {base**exp}")

base = int(input("Enter your base integer :"))
exp = int(input("Enter your exponential integer: "))
print(exponential(base,exp))

Youssef Hmini says

September 7, 2020 at 2:24 am

Love it! Especially for beginners like me!

Reply

el empotrador says

September 2, 2020 at 12:08 am

jhere I share my solutions with you, hope this helps.
question 1:

print("to play this game you have to introduce two numbers.")
number1 = input("please introduce the first number: ")
number2 = input("please introduce the second number: ")
product = int(number1) * int(number2)
if product >= 1000:
    sum = int(number1) + int(number2)
    print(sum)
else:
    print(product)

question 2:

for elements in range(11):
    print("the current number is ", elements, "the previous number was: ", elements - 1 , "the sum is:  ", elements + elements - 1)

question 3:

letter = input("please introduce a word: ")
print(letter[0: len(letter):2])

question 4:

string = str(input("please introduce a word: "))
while True:
    n = input("please introduce a number shorter than the length of the shosen word: ")
    m = int(n)
    if n.isdigit() and int(n) > len(string):
        print("please introduce a shorter number than string`s length")
    elif n.isdigit() and int(n) < len(string):
        print(string[m:])
        break
print("thanks for playing")

question 5:

listOfNumbers = input("please introduce five numbers separated by space: ")
Numbers = listOfNumbers.split()#here i have leant how to accept a list as an input
if (Numbers[0] == Numbers[4]):
    print("Given the list ", Numbers)
    print("the result is True")
else:
    print("Given the list ", Numbers)
    print("the result is False")

question 6:

 def divisibility(List):
    for elements in List:
        if elements % 5 == 0:
            print(elements)
givenList = [10, 20, 33, 46, 55]
divisibility(givenList)

question 7:

string = input("please introduce a string in wich there is at least a repeated element: ")
print("the shosen string is: ", string)
elementToCount = input("please introduce the element you want to count: ")
count = string.count(elementToCount)#to count how many time a element is repeated in a string I use the .count() mehtod. the estructure of this method is: namelist.count(the element we want to count)
print(elementToCount, "appears ", count, "times in the string")

question 8:

for elements in range(1, 6):
    print(" ".join(str(elements) * elements))

question 9:

num1 = input("please introduce a number: ")
if num1 == num1[-1::-1]:
    print("the original number ", num1, "is the same as its inverse")
else:
    print("the original number is not equal to its inverse")

question 10:

list1 = [10, 20, 23, 11, 17] 
list2 = [13, 43, 24, 36, 12]
newlist1 = [ ] 
newlist2 = [ ] 
finallist = [ ]  
for elements in list1:
    if elements % 2 == 0:
        newlist1.append(elements)#in this if statement I append the odd and even elements to two different list*
for e in list2:
    if e % 2 != 0:
        newlist2.append(e)
finallist = newlist1 + newlist2 #here I add the two different lists*
print(finallist)

question 11:

 number = input("please a list of numbers separated by space: ")
givennumber = number.split()
invertednumber = givennumber[-1::-1]
separatednumber = " ".join(invertednumber)
print(separatednumber)

question 12:

 totalIncome = int(input("please introduce your total income: "))
first = int(10000) * float(0.0)#when writing floats i have to use a dot instead of a comma 
second = int(10000) * float(0.1)
remain = totalIncome - int(20000)
final = remain * float(0.2)
totalTax = first + second + final
print("$1000 * 0 % + $1000 * 0,1 % + $ ",remain, " * 0,2 = $",totalTax)

question 15:

while True:
    base = input("please introduce the base of the power: ")
    if not base.isdigit():
        print("please introduce a number")
    else:
        break
while True:
    exp = input("please introduce the exponent: ")
    if not exp.isdigit():
        print("please introduce a number")
    else:
        break
def exponent(bs,ex):
    power = int(bs)**int(ex)
    return power
print(exponent(base,exp))

bhavya says

August 27, 2020 at 11:02 pm
```
a = int(input("Input an integer : "))
n1 = int( "%s" % a )
n2 = int( "%s%s" % (a,a) )
n3 = int( "%s%s%s" % (a,a,a) )
print (n1+n2+n3)
```
I saw this somewhere… I’m stuck here.. what is this % s
In an earlier code, I saw %e and %i…
please if anyone can explain, I’d be thankful
Reply
- Rakesh Singh Choudhary says
  
  August 31, 2020 at 9:00 pm
  1 .
```
a = int(input("Input an integer: "))
```
  Input method converts the user input to a string and then returns it, hence ‘int’ is prefixed here so that we can perform addition operation at the end.
  
  2.
```
n1 = int("%s" %a)
```
  The modulo % is referred to as a “string formatting operator”, here s stands for string representation. Here n1 = 4 if a is 4. You can run type(n1) and see it is str without int being prefixed.
  3.
```
n2 = int( “%s%s” % (a,a) )
```
  Similarly, here n2 = 44 if a is 4
  4.
```
n3 = int( “%s%s%s” % (a,a,a) )
```
  Again, n3 = 444 if a is 4
  
  Hence, n1 +n2 + n3 = 492
  Also, % e and % i are exponential and integer representations respectively. You can play around by putting these in print method.
  Reply

el empotrador says

August 26, 2020 at 6:14 pm

I could not solve exercise two. However, now i found what i consider the easiest solution:

for elements in range(11):
    print("the current number is ", elements, "the previous number was: ", elements - 1, "the sum is: ", elements + elements - 1)

serpilozdemir says

August 22, 2020 at 5:34 pm

Exercises are very excited for me thx
question 15 :

base=6
exponent=2
print('6 raises to power of 2 is', pow(base,exponent))

El empotrador says

August 25, 2020 at 11:19 am

Hello everyone,. I am new at learning python and I’ve done the first exercise in a different way. Can someone please tell me if it is correct or not??

 print("please introduce two numbers")
number1 = input("the first number you want to introduce is:  ")

number2 = input("the second number you want to introduce is: ")

product = int(number1) * int(number2)
if product > 1000:
	sum = int(number1) + int(number2)
	print("the sum of number1 and number2 is: ", sum)
else:
	print("the product of numer1 and number2 is: ", product)

Soumya says

August 21, 2020 at 9:39 pm

for question 15:

b=int(input("enter base"))
 e=int(input("enter exp"))
print(b,"raise to", e, "is",b**e)

DMS KARUNARATNE says

August 17, 2020 at 4:36 pm

Question No 9: another way to do it

def rever(num):
    #print (str(num))
    revnum=("".join(reversed(str(num))))
    #print (revnum)
    if str(num)==str(revnum):
        return True
    else:
        return False

Huirong Ai says

October 13, 2020 at 9:46 am

Mine is similar:

  def reverseCheck(number):
    print("original number:", number)
    new_number = int(str(number)[::-1])
    print("reversed number:", new_number)
    
    if number == new_number:
        print("The original and reverse number is the same")
    else:
        print("The original and reverse number is not the same")

corleone1986 says

June 27, 2022 at 11:26 pm

def checking(num):
    x = str(num)
    y = x [::-1]
    if x == y :
        print("yes")
    else:
        print("No")

checking(7658)

Dev says

August 14, 2020 at 11:49 pm

hey guys!!! Here’s a program from which u can code a calculator:

print("what operation do u want")
operator = input("Enter either +, -, * or /: ")

num1 = int(input("Enter first number: "))
num2 = int(input("Enter second number: "))

if operator == '+':
  print(num1, operator, num2, "=", num1+num2)
elif operator == '-':
  print(num1, operator, num2, "=", num1-num2)
elif operator == '*':
  print(num1, operator, num2, "=", num1*num2)
elif operator == '/':
  print(num1, operator, num2, "=", num1/num2) 
else:
  print("Invalid Operator")

Kundan Chaudhari says

August 20, 2020 at 11:10 pm

what if I want more than 2 values operations

Reply

Dev says

August 14, 2020 at 11:15 pm
Ques. 15:
```
def exponent(base, exp):
    result = base**exp
    print(base, "raises to the power ", exp, "is: ", result)

(exponent(5, 4))
```
Reply
- Dev says
  
  August 14, 2020 at 11:16 pm
  
  I hope this would be helpful to others!!!!
  Stay Home!!
  Stay Safe!!!!
  
  Reply
- Arkaprava Biswas says
  
  August 19, 2020 at 3:34 am
  
  best way
  
  Reply

Anusha says

August 6, 2020 at 10:53 am

question 15:

def exponent(base,exp):
    return base**exp
print(exponent(5,4))

Kristofer Kangro says

July 28, 2020 at 3:11 pm

Was its purpose to make solutions as difficult as possible simply for learning new ways of writing a simple program?
Some of the questions solutions were very challenging for me to understand as a beginner but I think it has a good and bad part to it. Bad side to it might be overcomplicating a simple question but on the other end it will teach you new ways to write a simple program.
In conclusion, questions were great and on-the-level for a beginner. Thank you!

Reply
- Vishal says
  
  July 29, 2020 at 11:13 am
  
  Thank you Kristofer for your valuable feedback.
  
  Reply
- Filippi says
  
  August 12, 2020 at 8:48 pm
  
  Your point is totally nonsense. The goal here is not to solve this problem and print an output, but study and learn python. The answares MUST be as difficult as possible, as this represents a guide were you can follow to learn a little bit, for free by the way.
  Vishal: Your job here was perfect, keep doing your best to put the most difficult answare you can get, thank you very much, best regarts
  
  Reply

Harshit Rautela says

July 26, 2020 at 1:00 pm

Q15 simple solution

def exponent(base, exp):
    result = base**exp
    print(result)


print(exponent(5, 4))

HARSH says

July 23, 2020 at 4:53 pm
```
#SHORTEST WAY OR QUESTION 9

number=input("ENTER THE NUMBER :")

if number==number[::-1]:
    print("PALINDROME")
else:
    print("NOT A PALINDROME")
```
Reply
- charlot fx says
  
  September 6, 2020 at 9:55 pm
  
  yeah i did the same way
  
  Reply
- Dushyant says
  
  October 13, 2020 at 8:36 pm
  
  This code is reversing a string and hence is inaccurate since any input taken in python without other function is always a string. You need to write different code for reversing a number which will be type integer
  
  Reply

Harshit Rautela says

July 22, 2020 at 1:18 pm

Q3 simple solution:


 def display_char(name):
    for i in range(0, len(name)):
        while i < len(name):
            print(name[i])
            i += 2
        break

print(display_char("pynative"))

Dev says

August 14, 2020 at 11:27 pm

here’s a solutio for Ques:3:

list1 = ["p","y","n","a","t","i","v","e"]
list2 = list1[0:8:2]

for str in list2:
  print(str)

Er. Tushar says

July 14, 2020 at 2:01 am

#Q15 Simple Solution


def exponent(base,exp):
    out=base**exp
    return print(out)

print("2 raise to the power 5 is:")
exponent(2,5)
print("-------------------------------------------")
print("5 raise to the power 4 is:")
exponent(5,4)

Nilotpal Shanu says

July 10, 2020 at 12:29 am
Q no.4
```
Str = "pynative"
host = str[4: ]
print(host)
```
Reply
- Rithesh says
  
  July 11, 2020 at 11:32 am
```
tive
```
  Reply

Aka code_teen says

July 6, 2020 at 8:20 pm

# Q2 Simple Solution :


prev_num = 0
for cur_num in range(10):
     print(f" current num is {cur_num} prev num is {prev_num} Sum: {cur_num+prev_num}")
     prev_num = cur_num

Good Luck! 😉

Alex says

July 5, 2020 at 7:18 am

My daughter studying from your tutorials; it is great!
please continue adding more exercises.

Vishal says

July 6, 2020 at 7:41 am

Thank you, Alex.

Tuur Six says

July 6, 2020 at 7:47 pm

oefening 3:


Has a little mistake if you type in a word that is even you don't have the last letter:
It should be:
def letter(let):
    for index in range(0,len(let)-0,2):
        print("index[",index, "] " + let[index])
inputStr = "can"
letter(inputStr)

Alimul Nishat says

July 4, 2020 at 11:01 pm
Hello Vikash bro,
I am a beginner of python programming language and found this platform best for solving problems for beginners.

And i found an alternative, short and easiest way to solve the problem no 14 is :
```
for i in range(5,0,-1):
    print(i*" *")
```
Reply

Khalid says

June 30, 2020 at 9:33 pm

My solution for number 9


def reverse_check(original_num):
    """"Reverses a given number and compares it with the original"""
    original_nums = str(original_num)
    num_len = len(original_nums)
    reverse_num = ""

    for num in range(num_len):
        reverse_num += original_nums[num_len - num - 1]

    if int(original_num) == int(reverse_num):
        print(f"original number {original_num}\nThe original and reverse number are the same: True")
    else:
        print(f"original number {original_num}\nThe original and reverse number are the same: False")


reverse_check(1221)

Khalid says

June 30, 2020 at 9:31 pm

My code for number9


def reverse_check(original_num):
    """"Reverses a given number and compares it with the original"""
    original_nums = str(original_num)
    num_len = len(original_nums)
    reverse_num = ""

    for num in range(num_len):
        reverse_num += original_nums[num_len - num - 1]

    if int(original_num) == int(reverse_num):
        print(f"original number {original_num}\nThe original and reverse number are the same: True")
    else:
        print(f"original number {original_num}\nThe original and reverse number are the same: False")


reverse_check(1221)

Houari_Aouinti says

June 27, 2020 at 4:55 am

Exercice 01:


count = lambda a, b: a * b if a * b < 1000 else a + b
print(count(25, 40))

Exercice 02:


def up(pNum, cNum, i):
    for n in range(i):
        print("Current Number:", cNum, "Previous Number:", pNum, "Sum:", cNum + pNum)
        pNum, cNum = cNum, cNum + 1
up(0, 0, 10)

Exercice 03:


def pEven(string):
    for m in range(len(string)):
        if m % 2 == 0: print(f"index {m} {string[m]}")
pEven("pynative")

Exercice 04:


split = lambda string, n: string[n:]
print(split("pynative", 4))

Exercice 05:


arr = [10, 20, 30, 40, 10]
firstLast = lambda arr: True if arr[0] == arr[-1] else False
print(firstLast(arr))

Exercice 06:


array = [10, 20, 33, 46, 55]
def arrayDvd(array):
    for i in array:
        if i % 5 == 0: print(i)
arrayDvd(array)

Exercice 07:


string = "Emma is good developer. Emma is a writer"
countString = lambda string: print(f"Emma appeared {string.count('Emma')} times")
countString(string)

Exercice 08:


def pattern(n, i):
    while n > i:
        print(f"{str(i) * i}")
        i +=1
pattern(10, 1)

Exercice 09:


rev = lambda number: True if number == number[::-1] else False
print(rev(str(121)))

Exercice 10:


fList, sList, rList = [10, 20, 23, 11, 17], [13, 43, 24, 36, 12], []
def oddEve(fl, sl):
    for i, j in zip(fList, sList):
        if i % 2 != 0 or j % 2 == 0: rList.append(i); rList.append(j)
    print(rList)
oddEve(fList, sList)

karan says

August 30, 2020 at 6:45 pm

wowww………………!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1

Reply

Daniel says

June 25, 2020 at 1:40 am

Hello,

I just started learning how to program. Well, just learning one program, Python. Now, I’m completely new to this stuff and I feel overwhelmed sometimes. Can anyone share some very easy easy problems that I can work on?

Thank you guys!

Reply
- Vishal says
  
  June 25, 2020 at 3:13 pm
  
  you’re welcome, Daniel
  
  Reply
Son_Volam says

June 23, 2020 at 5:44 pm
#Q9:
```
num1 = input()

if num1 == num1[-1::-1]:
    print("TRUE")
else:
    print("FALSE")
```
Reply
- HASAN says
  
  July 21, 2020 at 10:56 am
  Q(9).
```
x=input('Enter a number: ')
if x[::-1]==x:
	print(True)
else:
	print(False)
```
  Reply
- el empotrador says
  
  August 31, 2020 at 6:23 pm
  
  thanks for publishing your answer, it was very helpful for me
  
  Reply

Son_Volam says

June 23, 2020 at 5:44 pm

#Q9:


num1 = input()

if num1 == num1[-1::-1]:
    print("TRUE")
else:
    print("FALSE")

Steph says

June 17, 2020 at 7:43 pm
Regarding problem 9, is it bad practice or something to convert an int to a string to reverse it and then convert it back? This process just seems more straight forward than the answer provided.

Here is what I came up with(includes some of the answer):
```
def reverseCheck(org_number):
  test_string = str(org_number)
  reverse_str = test_string[::-1]
  if int(reverse_str) == org_number:
    print("True")
  else: 
    print("False")
  print(int(reverse_str))

print("Is the original and reverse number the same?", reverseCheck(121))
```
Reply

thet wai htut says

June 17, 2020 at 11:55 am

question 8


for i in range(1,6):
    for j in range(1,i+1):
        print(i,end='')
    print()

Dhyan says

June 13, 2020 at 4:04 pm

Hey buddy doing a wonderful job by posting such useful exercise for practice 🙂 keep it up

Reply
- Vishal says
  
  June 15, 2020 at 5:08 pm
  
  Thank you, Dhyan.
  
  Reply
Nikita says

June 9, 2020 at 8:26 pm

Please bring back the old code editor 🙁

Reply
- Vishal says
  
  June 10, 2020 at 12:21 am
  
  Hey Nikita, Please let us know if you are facing any problem. Thank you
  
  Reply
  - Dev says
    
    August 14, 2020 at 11:34 pm
    
    Sir, I only knew about the only theory about Python but when I explore on chrome for some exercises I selected this…. and this has improved my coding skills!! God Bless You
    Stay Home!!!!
    Stay Safe!!!!
    
    Reply
narender says

June 8, 2020 at 12:01 pm
Question 3:
solution :
```
s = input(‘Enter your string : ‘)
L = len(s)
print(s[0:L:2])
```
Reply
- Jon Blackwell says
  
  June 24, 2020 at 3:51 am
  
  You don’t need line 2, and variable “L” in line 3 for your solution.
  
  Reply
Ray says

June 7, 2020 at 11:20 pm

Thank you very much for the exercises. They are extremely helpful for a beginner.
Keep up the good work

Reply
- Vishal says
  
  June 8, 2020 at 9:26 am
  
  You are welcome, Ray
  
  Reply
Logan Peterson says

June 7, 2020 at 11:28 am
For the Emma question, I have a short solution that works very well.
```
a = 'Emma is a good developer. Emma is a good writer'
print(a.count('Emma'))
```
Reply

Angshumaan Basumatary says

June 6, 2020 at 5:35 pm

Question 10: Tried using user input

def calculate(a, h):
    k = [z for z in a if z % 2 != 0]
    b = [y for y in h if y % 2 == 0]
    c = k + b
    return c


a = int(input("Enter the size: "))
angshu = [int(x) for x in input("Enter first list").strip().split()][:a]
basu = [int(x) for x in input("Enter second list").strip().split()][:a]
i = calculate(angshu, basu)
print(i)

John says

June 3, 2020 at 4:11 pm

#Question 9 alt. solution

def isrev(x):
  x1 = list(str(x))
  x2 = list(str(x))
  x2.reverse()
  print(x1)
  print(x2)
  print(x1==x2)
  
isrev(121)

Steven-r says

May 23, 2020 at 5:29 am

Hi Vishal
As a newbie, your exercises come in really helpful and useful. Thanks so much for posting! much appreciated.

Reply
- Vishal says
  
  May 26, 2020 at 7:48 pm
  
  You are welcome, Steven-r
  
  Reply

jyotirmoy ghose says

May 23, 2020 at 12:17 am

Question 1:

def mul_or_sum(num1,num2):
   
    output= num1*num2
    if(output>1000):
        value = num1 + num2
        print("the result is : ", value)
    else:
        print("the result is : ", output)

num1= int(input("enter num1: "))
num2=  int(input("enter num2: "))
mul_or_sum(num1,num2)

Question 2:

def current_previous(number):
    for i in range(1,number):
         print("current number :", i,"previosu number: ", i-1,"sum: " ,i+ i-1)

number= int(input("enter the numebr: " ))
current_previous(number)

Question 3:

def display_even_number_string(st):
    lent=len(st)

    for i in range(0,lent,2):
        print("index[", i ,"]", st[i])


st= str(input("enter string : "))
print("original string:",st)
display_even_number_string(st)

Question 4:

def remove_string_char(st):
    
    New_st= st[number ::]
    print("remaining strings are ---> :",str(New_st))

st = str(input("enter string : ")) 
st_len=len(st)
print("length of the string is :",st_len)
number= int(input("enter the number < string len : "))
remove_string_char(st)

Question 5:

def list_check(list):
     if(list[0]==list[-1]):
         print("result is: ",bool(1))
     else:
         print("result is: ",bool(0))    

list=[]
number = int(input("enter number : "))
for i in range(number):
    value=int(input("entered number to the list : "))
    list.append(value)
    print("after adding :",list)
list_check(list)

Question 6:

list=[]
number = int(input("enter number : "))
for i in range(number):
    value=int(input("entered number to the list : "))
    list.append(value)
    print("after adding :",list)

for i in range(number):
    if(list[i]%5==0):
        print(list[i],end=" ")
    else:
        print("nothing found")

Question 7:

st=str(input("enter your string: "))
def word_count(st):
    count = dict()
    words = st.split()
    print(words)
    for i in words:
        if i in count:
            count[i] += 1
        else:
            count[i] = 1
    return count
print( word_count(st),end=" ")

Question 8: Print the following pattern

for i in range(1,6):
    for j in range(i):
        print(i,end=" ")
    print("\n")

Question 9:

number = 12
reverse=0
while(number>0):
     remainder = number%10
     print(remainder)
     reverse = (reverse*10)+remainder
    
     number=number//10
     
print(reverse)

Question 10:

list1=[10, 20, 23, 11, 17]
list2=[13, 43, 24, 36, 12]
value1= len(list1)
value2=len(list2)
new_list1=[]
new_list2=[]
final_list=[]
for i in list1:
    if(i%2!=0):
        new_list1.append(i)

for j in list2:
    if(j%2==0):
        new_list2.append(j)    

final_list=new_list1+new_list2
print(new_list1)    
print(new_list2)  
print(final_list)

jyotirmoy ghose says

May 23, 2020 at 12:16 am

list1=[10, 20, 23, 11, 17]
list2=[13, 43, 24, 36, 12]
value1= len(list1)
value2=len(list2)
new_list1=[]
new_list2=[]
final_list=[]
for i in list1:
    if(i%2!=0):
        new_list1.append(i)

for j in list2:
    if(j%2==0):
        new_list2.append(j)    

final_list=new_list1+new_list2
print(new_list1)    
print(new_list2)  
print(final_list)

Sandeep says

May 22, 2020 at 11:22 pm

Alternative way for question 1: **please excuse for any mistakes

 
def multiplication_or_sum(num1, num2):
    if (num1 * num2) > 1000:return('The result is %s' %(num1+num2))
    else:return('The result is %s' %(num1*num2))

num1 = int(input('enter first number:\n'))
num2 = int(input('enter second number:\n'))
print(multiplication_or_sum(num1, num2))

keshav ranjithlall says

May 7, 2020 at 10:06 pm

hey bro for solution 2: what is the + 4 in the [i : 1 + 4] for?

Reply

Mr.HAZANI says

April 20, 2020 at 7:46 pm

One point to clear:
I found that the comment section ignored the spaces that I put behind every line and changed a few marks.
Paste this modified version in your Python to get it working:

#Principal Variable
number = 121

#Function
def ReverseNum (num):
    snum = str(num)
    reversed = snum [::-1]
    Reversed = int(reversed)
    return Reversed

#Output :

print("original number", number)
if number == ReverseNum(number) :
    print("The original and reverse number is the same:", True)
else:
    print("The original and reverse number is the same:", False)

Oury says

April 27, 2020 at 12:38 pm

def same_backwards(num):
    '''returns true if num is the same when written backwards'''
    return str(num) == str(num)[::-1]

Santosh says

April 19, 2020 at 10:29 am

Last Question :

list1 = [1,2,4,6,2,3]
list2 = [3,432,433,22,442]

list3 = [x for x in list1 if x % 2 != 0] + [x for x in list2 if x % 2 == 0]
print(list3)

Vishal says

April 19, 2020 at 8:31 pm

Thank you, Santosh for an alternative solution

Reply
- Ankana says
  
  April 25, 2020 at 8:26 am
```
list1=[10,20,40,50,70]
list2=[20,60,80,90]

list3=list1[1::2]+list2[1::2]
print(list3)
```
  Reply

cyberhoax says

April 15, 2020 at 5:13 pm
A7. A very useful and short way to count the words in a string
```
a = "Emma is good developer. Emma is a writer"
c = a.count("Emma")
print(c)
```
Reply
- Vishal says
  
  April 15, 2020 at 9:22 pm
  
  Hey cyberhoax, Thank you for an alternative solution.
  
  Reply
- Nikhil K says
  
  April 27, 2020 at 10:18 pm
```
a = "Emma is good developer. Emma is a writer"
print(a.count("Emma"))
```
  Reply
- mahesh says
  
  May 11, 2020 at 7:54 pm
```
a = "Emma is good developer. Emma is a writer"
c = a.count(input('enter the word: '))
print(c)
```
  Reply
saba nadiradze says

April 3, 2020 at 12:22 am

I am learning coding for 2 weeks and it totally impoved my skills. thank you for your exercises.

Reply
- Vishal says
  
  April 3, 2020 at 8:31 pm
  
  Hey Saba, Thank you for your kind words.
  
  Reply
  - Sidra says
    
    October 1, 2020 at 5:43 pm
    
    I need the answer of two questions can I plzz help me
    
    Reply
- Nikhil muppidi says
  
  April 9, 2020 at 11:25 pm
  My code for question no 10:
```
mylist1 =  [10, 20, 23, 11, 17]
mylist2 =  [13, 43, 24, 36, 12]
oddlist = [i for i in mylist1 if i%2!=0]
evenlist = [i for i in mylist2 if i%2==0]
print(oddlist+evenlist)
```
  Reply
  - Akbar says
    
    April 18, 2020 at 7:02 am
    
    best way
    
    Reply
Anca says

April 2, 2020 at 9:40 pm

Hello! Can someone explain to me the solution for example 7?
I dont understand what this line does:
count += statement[i:i + 4] == ‘Emma’. I know about slicing but that +4 confuses me. thanx

Reply
- Vishal says
  
  April 3, 2020 at 8:44 pm
  
  Hi Anca,
  
  The statement is in the loop. Slicing starts with the start index (included) and ends at end index (excluded)
  
  For example, in the first iteration of a loop i =0
  
  Here statement[ 0 : 0 + 4] i.e., statement[0 : 4] and it will return “Emma” and if it is equal to “Emma”, then count will increment as statement become true and evaluate to value 1
  
  Reply
Florian Fasmeyer says

March 14, 2020 at 8:01 pm
Alternative solutions to pynative.com – basic exercise for beginners.
# Q1.
```
a, b = int(input('Input a')), int(input('Input b'))
prod = a*b
print(prod if prod < 1000 else a+b)
```
# Q2.
```
nums = range(10)
prev = 0
for curr in nums:
  print(prev + curr)
  prev=curr
```
# Q3.
```
usr_str = input('Feed me a string!')
print(usr_str[0::2], sep='\n')
```
# Q4.
```
str, n = 'pynative', 4
print(str[n:])
```
# Q5.
```
nums = [1, 2, 3, 4, 5, 1]
print(nums[0] == nums[-1])
```
# Q6.
```
nums = range(0, 60)
print(*[num for num in nums if num%5 == 0], sep='\n')
```
''' Take-aways

Q1 – When a condition is simple, use a ternary op.: x if (cond) else y.
Q2 – If you are using indexers, you are doing it wrong!
Q3 – Don't iterate lists, use accessors instead: [start: stop: step]
Q4 – Please, use accessors instead! ref.Q3
Q5 – DO NOT IF/ELSE a condition if you return booleans! YOU IMBECILE!
Q6 – * operator is used to unpack: a, b, c = *[1, 2, 3]. You'll love it!

Q+ – You gotta learn list comprehensions: [exp(item) for item in list].
Q+ – List comprehensions with filters: [item for item in list if cond(item)]
Reply
- Florian Fasmeyer says
  
  March 14, 2020 at 8:05 pm
  
  As I can see you did not expect me to include tags!
  Too bad, I can’t change it now. 🙂
  
  Reply
  - Vishal says
    
    March 20, 2020 at 7:03 pm
    
    Thank you, Florian Fasmeyer. Now onwards Please add your code between <pre> </pre> tag for correct indentation.
    
    Reply
- Vishal says
  
  March 20, 2020 at 6:58 pm
  
  Thank you Florian Fasmeyer for alternative solutions
  
  Reply
- newbie noob says
  
  March 26, 2020 at 3:04 pm
  
  Hey. I have no idea about any of this. But on Q1 with the answer 1000?
  
  I’m so new to all of this I didn’t quite understand why the less than symbol is being used rather than the more than symbol. Thank you in advance
  
  Reply
  - Vishal says
    
    March 26, 2020 at 7:01 pm
    
    Hi newbie, I have updated the answer. In your case, the result should be 1000.
    
    Reply
- jeena gurung says
  
  June 9, 2020 at 7:05 pm
  
  Thank you so much for the alternative solutions. Being a newbie to Python, I was struggling understand author’s solutions, you made it lot easier.
  Thank you so much again.
  
  Reply

Junaid Ali says

February 7, 2020 at 10:45 pm

My Practice :

#Question 1: Accept two int values from the user and return their product.
# If the product is greater than 1000, then return their sum

m=int(input("Enter Number: "))
n=int(input("Enter Number: "))
r=m*n
if r>1000:
    print(m+n)
else:
    print(r)

#Question 2: Given a range of numbers.
# Iterate from o^th number to the end number and print the sum of the current number and previous number.

n=int(input("Enter Number for fibonnica series : "))
a=0
b=1
for i in range(n):
    if i == 0:
        print("0")
    else:
        c=a+b
        a=b
        b=c
        print(c)

#Question 3: Accept string from the user
# and display only those characters which are present at an even index

str = "pynative"
for i in range(len(str)):
    if i%2==0:
        print(str[i])

#Question 4: Given a string and an int n,
# remove characters from a string starting from zero up to n and return a new string

def remove(n):
    return str[n:]

str=input("Enter string : ")
n=int(input("Enter Index : "))
print(remove(n))

#Question 5: Given a list of ints, return True if first and last number of a list is same

n=int(input("How many numbers you will enter into list : "))
A=[]
for i in range(n):
    a=int(input("Enter Number : "))
    A.append(a)

if A[0]==A[-1]:
    print("First and last number is same")

#Question 6: Given a list of numbers,
# Iterate it and print only those numbers which are divisible of 5

n=int(input("How many numbers you will enter into list : "))
A=[]
for i in range(n):
    a=int(input("Enter Number : "))
    A.append(a)

for j in A:
    if j%5==0:
        print(j,end=" ")

#Question 7: Return the number of times that the string appears anywhere in the given another string

n=input("Enter phrase that repeats the words : ")
r=input("Enter repeated word : ")
c=0
n=n.split()
for i in n:
    if i in r:
        c+=1
print(c)

#Question 8: Print the following pattern

n=int(input("Enter Number of rows :"))
for i in range(1,n+1):
    for j in range(1,n+1):
        if i>j or i==j:
            print(i,end="")
        else:
            print(end="")
    print()

#Question 9: Reverse a given number and return true
# if it is the same as the original number

n=input("ENter Number : ")
if n==n[::-1]:
    print(n)

#Question 10: Given a two list of ints create a third list
# such that should contain only odd numbers from the first
# list and even numbers from the second list
#Merged List is [23, 11, 17, 24, 36, 12]

A = [10, 20, 23, 11, 17]
B = [13, 43, 24, 36, 12]
C=[]
for i in A:
    if i%2==1:
        C.append(i)
for j in B:
    if j%2==0:
        C.append(j)
print(C)

AMEENGT3 says

February 27, 2020 at 10:25 pm

NICE WORK

Reply
harpoon says

February 28, 2020 at 9:21 am

thank u this helped me a lot

Reply
- Vishal says
  
  March 26, 2020 at 7:02 pm
  
  I am glad it helped you.
  
  Reply

AMEENGT3 says

March 6, 2020 at 11:26 am

import turtle 
import time 
import random 
  
print ("This program draws shapes based on the number you enter in a uniform pattern.") 
num_str = input("Enter the side number of the shape you want to draw") 
turtle.speed(999)
if num_str.isdigit(): 
    squares = int(num_str) 
  
angle = 180 - 180*(squares-2)/squares 
  
turtle.up 
  
x = 0 
y = 0
turtle.setpos(x, y) 
  
  
numshapes = 9
for x in range(numshapes): 
    turtle.color(random.random(), random.random(), random.random()) 
    x += 5
    y += 5
    turtle.forward(x) 
    turtle.left(y) 
    for i in range(squares): 
        turtle.begin_fill() 
        turtle.down() 
        turtle.forward(40) 
        turtle.left(angle) 
        turtle.forward(40) 
        print (turtle.pos()) 
        turtle.up() 
        turtle.end_fill() 
  
time.sleep(11) 
turtle.bye()

AMEENGT3 says

March 6, 2020 at 11:29 am

Anyone can try this

Reply

Vishal says

March 9, 2020 at 6:55 pm

Great Junaid Ali. Keep it up

Reply

Houssem says

September 6, 2021 at 4:04 pm

Try this one:

def makeList(lst1, lst2):
    f_lst = list()
    for i,a in zip(lst1, lst2):
        if i%2 != 0: f_lst.append(i)
        if a%2 == 0: f_lst.append(a)
    return f_lst

Cero says

February 7, 2020 at 4:08 pm

class Dijak:
  def_init_(self, imeInPriimek,odelek,stDosezenihTock,vrstaOcene):
    self.imeInPriimek=imeInPriimek
    self.odelek=odelek
    self.stDosezenihTock=stDosezenihTock
    self.vrstaOcene=vrstaOcene
  def vrniOceno(self):
    odstotek=self.stDosezenihTock/28*100
    if(odstotek>=88):
      return(5)
    elif(odstotek>=75):
      return(4)
    elif(odstotek>=63):
      return(3)
    elif(odstotek>=50):
      return(2)
    else:
      return(1)
  def vrniImeInPriimek(self):
    loci=self.imeInPriimek.split("")
    niz=loci[1]+","+loci[0]
    return(niz)
  def izpisObjekta(self):
    print(self.vrniImeInPriimek()+"("+self.odelek+")"+"ocena :"+str(self.vrniOceno)+","+self.vrstaOcene)
  def shraniPodatek(N):
    for i in range(N)
      objekt=Dijak(input("Dijak :")),input("Odelek :"),int(input("Tocke :")),input("Vrsta :"))
      seznam.append(objekt)
    return(seznam)
  def VrniOceneDijaka(seznamdijakov,imeInPriimek):
    for s in seznamdijakov:
      if(s.imeInPriimek==imeInPriimek):
        s.izpisObjekta()
  def povprečjeRazreda(seznamdijakov,odelek,vrstaOcene):
    vsota=0
    stevec=0
    for s in seznamdijakov:
      if(s.vrstaOcene==vrstaOcene and s.odelek==odelek):
        vsota+=s.vrniOceno()
          stevec+=1
    print(vsota/stevec)

Grant says

February 2, 2020 at 8:45 am

This is more simplistic code for #5:

l = []
question = str(input("Do you want to enter a number; enter yes or no: "))

while question == 'yes':
    numbers = int(input("Enter a number here: "))
    question = str(input("Do you want to continue entering numbers; enter yes or no: "))
    l.append(numbers)

if l[0] == l[-1]:
    print("The first and last numbers of", l , "are the same!")
else:
    print("The first and last numbers of", l , "are not the same!")

Vishal says

February 3, 2020 at 11:05 pm

Grant, Thank you for your solution

Reply

magesh says

January 29, 2020 at 7:20 pm
Question 9: Reverse a given number and return true if it is the same as the original number
```
print(True) if(user_input == user_input[::-1]) else print(False)
```
Reply
- Vishal says
  
  March 26, 2020 at 7:04 pm
  
  Thank you for alternative solutions
  
  Reply
magesh says

January 29, 2020 at 7:04 pm
Another easiest solution for Question 8
```
for i in range(1,6):
    print(" ".join(str(i)*i),end="\n")
```
Reply
Eric says

January 7, 2020 at 10:18 pm

Hey,

As a beginner Python programmer, these questions are extremely helpful. Even when I can’t solve them on my own and peek at the given solutions, I make sure that I 100% understand the solutions before moving on to the next question. Thanks for your hard work.

Regards,
Eric

Reply
- Vishal says
  
  January 8, 2020 at 7:34 pm
  
  Eric, Thank you for your appreciation
  
  Reply
Kirich says

November 25, 2019 at 8:02 pm
Another solution for question 7:
```
test_str = "Emma is a good developer. Emma is also a writer"
counter = test_str.count('Emma')
print("Emma appeared " + str(counter) + '  times')
```
Does it count?
Reply
- Vishal says
  
  December 4, 2019 at 3:54 pm
  
  Yes. The output is Emma appeared 2 times
  
  Reply
sreekanth says

November 4, 2019 at 6:03 pm

This is a very nice practice for the beginners who wanted to kick off their python programming .

Reply
- Vishal says
  
  November 5, 2019 at 8:56 am
  
  Sreekanth, I am glad it helped you.
  
  Reply
  - Kenneth Jay Paradiang says
    
    December 4, 2019 at 11:44 am
    
    example shows parenthesis on if statements, but pycharm pose it as a pep8 violation. is it python 2 if statement structure?
    
    Reply
    - Vishal says
      
      December 4, 2019 at 3:49 pm
      
      All codes are developed using Python 3. You can remove parentheses
      
      Reply
Dwayne Lewis says

October 5, 2019 at 10:36 am

The admin of this site is doing a really good job and am really grateful for that. Also can the admin give suggestion of how to learn programming in python because i most of the basic stuff in Python but I’m still having problem putting the syntax together.

Reply
- Vishal says
  
  October 7, 2019 at 7:23 pm
  
  Thank you, Dwayne Lewis, Answer is practice. Try to solve as many examples as you can. or try to solve StackOverflow basic questions
  
  Reply
- AKASH says
  
  October 19, 2019 at 3:45 pm
  
  all these codes are in 3.7 or 2.7 ???
  
  Reply
  - Vishal says
    
    October 20, 2019 at 11:04 pm
    
    Python 3.7
    
    Reply
DevOps in Training says

September 29, 2019 at 2:43 am
For number 10 I used a list comprehension.
```
def mergeLists(list1, list2):
    list3 = list1 + list2
    mergedList = [x for x in list3 if x%2 != 0]
    return mergedList

mergeLists(listOne, listTwo)
[23, 11, 17, 13, 43]
```
Reply
- Zlatko says
  
  November 23, 2019 at 6:57 pm
  
  yeah, but you have only odd numbers…
  
  Reply

Mc Quest says

September 11, 2019 at 11:37 pm

Another possible solution for Q 1:

# program to accept two inputs and return sum

firstNo = int(input("Enter first number: "))
secondNo = int(input("Enter second number: "))
print(" ")
product = (firstNo * secondNo)
print(" ")

if product > 1000:
    print("The result is ", firstNo + secondNo)
else:
    print("The result is ", product)

Sumit says

September 1, 2019 at 9:31 pm

# Example 3

def example():
    inp_str = input("Enter on string : ")
    print("Your input string is :",inp_str)
    for i in range(0, (len(inp_str)-1),2):
        print("Index ",[i],":",inp_str[i])
example()

Ahmad says

August 10, 2019 at 6:34 pm
i think Answer to question 3 is as simple as this:
```
word = input('Please enter a word: ')

print(word[0 : : 2])
```
Reply
- Ahmad says
  
  August 10, 2019 at 6:37 pm
  in case of Even its:
```
print(word[ 1 :  : 2])
```
  Reply

Ahmad says

August 10, 2019 at 2:06 pm

The actual answer to the Question 1 should be:

result = int()

while result  1000:
        print('The result is: ', first_number + second_number)
        break

VIgnesh says

July 17, 2019 at 3:05 pm
My answer for question number 8
```
str1 ='012345'
for i in range(len(str1)):
    print(str1[i]*i)
```
Reply
Víctor says

July 12, 2019 at 11:25 pm
Hi, there is a small mistake in the solution of question 3. Specifically the error is in line 2:
for i in range(0, len(str)-1, 2).
As it is written, the last character of a word that has an odd number of characters, for example ‘tiger’ , will never be printed out, even if this last character is present at an even idex, like in this case the letter ‘r’. This is because the built-in function range() doesn’t include the last value of the specified range. For example, list(range(0,5)) will return [0, 1, 2, 3, 4], excluding the number 5.
Therefore, the correct solution is:
```
for i in range(0, len(str), 2) #remove the -1
for i in range(0, len(str)-1, 2):
```
Reply
Barry Ford says

July 4, 2019 at 5:06 am
On Question 8, I made my solution accept user input as to the height of the pyramid. See below
```
height = int(input("Height? "))
for num in range(height):
    for j in range(num + 1):
        print(num + 1, end=" ")
    print()
```
Reply

Zhixiang says

July 3, 2019 at 11:34 pm

on exercise 1:

a = int(input('Please enter the first integer: '))
b = int(input('Please enter the second integer: '))

c = a * b

print ('a * b =', c)

if c > 100: 
  d = a + b
  print('sum of a and b when d >100:', a + b)

Chris says

June 11, 2019 at 3:19 am
Here’s a shorter solution for question 3 🙂 It uses slicing with a step of 2.
```
str = "pynative"
"".join(str[::2])
```
Output:
‘pntv’
Reply

maxime says

June 7, 2019 at 5:49 pm

There is a short way to attend question 9.

# Reverse a given number and return true if it is the same as the original number
original_number = input("Enter the number: ")
reverse_number = original_number[::-1]
if int(original_number) == int(reverse_number):
    print("The reversed number is the same as the original one.")
else:
    print("They are not the same.")

Robert Carew says

May 16, 2019 at 8:39 pm

Hello Vishal.

I really am grateful for the work you put in to make this resource. This comment is a token of my thanks. I am a beginner programmer and this is helping me increase my knowledge of Python.

Cheers.

Reply
- Vishal says
  
  May 16, 2019 at 10:14 pm
  
  Hey Robert, I am glad you liked it
  
  Reply
  - Lucky says
    
    September 5, 2019 at 5:36 pm
    
    hi Vishal , i am from south africa i also need help
    
    Reply
    - Vishal says
      
      September 5, 2019 at 9:01 pm
      
      Hi Lucky, Please tell me
      
      Reply
Bill Hardwick says

May 16, 2019 at 3:41 pm

Just a general comment on these Comment boxes. Like me, I suspect most respondents post the python code by pasting from an editor (such as the Trinket one provided here). Unfortunately the mark-up behind this Comment box does not appear to support tab characters, so the all-important pythonic indentation is lost. This could be very confusing for a beginner, particularly where multi-line code blocks are concerned. Any way you could rectify this please?

(In the meanwhile, the workaround would appear to be to manually replace any tabs with 4 spaces, but that is prone to editing error or omissions.)

Reply
- Vishal says
  
  May 16, 2019 at 10:29 pm
  
  Hey, Bill
  
  You can paste your code in pre HTML tag so it can keep indentation as it is.
  
  Reply

Bill Hardwick says

May 16, 2019 at 3:30 pm

My alternative solution to Q10:

list1 = [1, 2, 3, 4, 5, 6, 7, 8, 9]
list2 = [21, 22, 23, 24, 25, 26]
list1_odd = [e for e in list1 if e%2 != 0]
list2_even = [e for e in list2 if e%2 == 0]
list3 = []
list3.extend(list1_odd)
list3.extend(list2_even)
print(list1)
print(list2)
print(list3)

Bill Hardwick says

May 16, 2019 at 3:27 pm

May I suggest an alternative solution to Q7.

txt = 'Emma is a good developer. Emma is also a writer.'
count = 0
lst = txt.split()
for word in lst:
  if word == 'Emma':
    count += 1
print(f'Emma appears {count} times in the string.')

Neeraj Raut says

July 13, 2019 at 9:35 pm

Just a little different and shorter approach if i may suggest…

def string(s):
    c=s.count("Emma")
    return f"Emma occurred {c} times"
print(string('Emma stone is great actor. Emma is genius. Emma play sports'))

Improve says

May 15, 2020 at 10:11 am

? at first try I’ve solved by using this approach ?

Reply

Richard says

April 23, 2019 at 9:44 pm

Hi Vishal
I think to help the beginners better, you should solve the problems without always using user defined functions, just Incase those that are not familiar with it.
Don’t mind my frequent responses, am in love with PYTHON that’s why.
Thanks

Reply
- Vishal says
  
  April 24, 2019 at 6:50 am
  
  Hey, Richard Thank You for your suggestions.
  
  Reply
Richard says

April 23, 2019 at 3:00 pm
Vishal I think your solution to question1 is a little verbose, a straight if and else statement does the job just well, it Will be much easier for beginners. I.e
```
if product>1000:
    print('sum is: ', sum)
else:
    print('product is: ', product)
```
Reply

Mdsambo says

March 30, 2019 at 7:39 pm

user_input = input("input any string") 
for a,  j in enumerate(range(0, len(user_input), 2)):
         print(user_name)

# means square bracket

Mdsambo says

March 30, 2019 at 7:42 pm
```
Print(username(j))
```
its a square bracket
Reply

Toby Robert says

March 14, 2019 at 12:35 pm

May I suggest a more concise solution to question no. 9?

orignum= 234565432
revnum=(str(orignum)[::-1])
print(revnum)
if str(orignum) == revnum:
   print('True: The number is the same when reversed')

Vishal says

March 14, 2019 at 12:50 pm

Thank you Toby for your valuable suggestion. It will help others also.

Reply

Jasmine_a says

October 24, 2020 at 6:10 pm

hi sorry, but can you check this code, please?
I want to write a program in which you get multi-digit numbers from the user without any restrictions on repetition and digits. It can be up to n digits and I want to add those digits in order so that I have only one digit in my output, like this example :
input = 123874 solving = 1+2+3+8+7+4 =25 , 25 = 7 output = 7
or , input = 14 output = 5

and this is my code but it doesn’t work. can you debug this in a way that it works and be able to give any input from user?

def sumdigits_nonrecursive(number):
    result = number
    while result > 10:
        ac = 0         # initialize a temporary accumulator
        for c in str(result):
            # for each character of the "result" string, add to accumulator the digit value
            ac = ac + int(c)
        result = ac
    return result

Richard_K says

February 15, 2022 at 7:35 pm

Hello Jasmine_a.
A little late, but this may help someone:


def sumdigits_nonrecursive(number):
    result = number
    while result > 10:
        ac = 0         # initialize a temporary accumulator
        for c in str(result):
            # for each character of the "result" string, add to accumulator the digit value
            ac = ac + int(c)
        result = ac
    return result

#use input() to allow user to enter the number
number = input("Enter number to evaluate: ")
#convert user input to int
number = int(number)
#call your function inside print() to display result returned
print( sumdigits_nonrecursive(number) )

seraph776 says

April 30, 2022 at 7:40 am



def sumdigits_nonrecursive(n: int) -> int:
    """sum digits non-recursively"""
    while True:
        n = sum([int(i) for i in str(n)])
        if len(str(n)) == 1:
            break
    return n

print(sumdigits_nonrecursive(123874)) # 7
print(sumdigits_nonrecursive(14)) # 5

Joseph says

March 8, 2019 at 11:35 am
Dear Vishal,
I tried Question 1 and failed as compare to your solution. I think your question was wrongly phrased.
My solution is
#It is assumed user will enter integers only, no check on input
```
first = int(input("Enter first integer "))
second = int(input("Enter second integer "))

product = first * second
print ("Product is ", product)

if product > 1000:
    print("Sum as product > 1000 : ", first + second)
```
Your Question should have been
Question 1: Accept two integer values from user.
Calculate their product and if the product is greater than 1000
then return their sum otherwise return their product.

Regards,
Joseph
Reply
- Ahmad says
  
  August 10, 2019 at 2:02 pm
  You are right, according to the question the solution should be:
```
result = int()
while result  1000:
        print('The result is: ', first_number + second_number)
        break
```
  In this way, It will keep asking you the numbers until the product is not greater than 1000 and If the Product is greater than 1000 then it will show you the result by adding both numbers
  Reply
- Ahmad says
  
  August 10, 2019 at 2:04 pm
  sorry the actual solution: some how skipped to copy some lines
```
result = int()

while result  1000:
        print('The result is: ', first_number + second_number)
        break
```
  Reply

Table of contents

Exercise 1. Arithmetic Product and Conditional Logic

Exercise 2. Cumulative Sum of a Range

Exercise 3. String Indexing and Even Slicing

Exercise 4. String Slicing and Substring Removal

Exercise 5. Variable Swapping (The In-Place Method)

Exercise 6. Calculating Factorial with a Loop

Exercise 7. List Manipulation: Add and Remove

Exercise 8. String Reversal

Exercise 9. Vowel Frequency Counter

Exercise 10. Finding Extremes (Min/Max) in a List

Exercise 11. Removing Duplicates from a List

Exercise 12. List Comparison and Boolean Logic

Exercise 13. Filtering Lists with Conditional Logic

Exercise 14. Substring Frequency Analysis

Exercise 15. Nested Loops for Pattern Generation

Exercise 16. Numerical Palindrome Check

Exercise 17. Merging Lists with Parity Filtering

Exercise 18. Integer Digit Extraction and Reversal

Exercise 19. Multi-Tiered Income Tax Calculation

Exercise 20. Nested Loops for Multiplication Tables

Exercise 21. Downward Half-Pyramid Pattern

Exercise 22. Custom Exponentiation Function

Exercise 23. Check Palindrome Number

Exercise 24. Generate Fibonacci Series

Exercise 25. Check Leap Year

Exercise 26. Merging Two Dictionaries

Exercise 27. Finding Common Elements (Intersections)

Exercise 28. Odd/Even List Splitter

Exercise 29. Word Length Analysis

Exercise 30. Word Frequency Counter (The Histogram)

Exercise 31. Print Alternate Prime Numbers

Exercise 32. Dictionary of Squares (Mapping Logic)

Exercise 33. Character Replacer (Data Sanitization)

Exercise 34. Print Reverse Number Pattern

Exercise 35. Digit Detection in Strings

Exercise 36. Capitalize First Letter (Title Case)

Exercise 37. Simple Countdown Timer

Exercise 38. File Creation and Basic I/O

Exercise 39. External File Word Counter

Exercise 40. Introduction to Classes (OOP)

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