i = 0
 
while i < 10:
    print("Hello")
    i += 1

Output:

Hello
Hello
Hello
Hello
Hello
Hello
Hello
Hello
Hello
Hello

Key points:

• We initialize i = 0 before the loop.
• The loop body prints and then increments i.
• When i reaches 10, the condition becomes false and the loop ends.

i = -1
 
while i > 0:
    print("Hello world!")

No text is printed because the condition i > 0 is false from the very beginning. A while loop may execute zero times.

if statement — executes once:

i = 0
if i < 10:
    print("Hello")

while statement — executes repeatedly:

i = 0
while i < 10:
    print("Hello")
    i += 1

Key takeaway: The if block runs at most once; the while block runs until the condition becomes False.

Using a for loop:

for i in range(101):
    print(i)

Using a while loop:

i = 0
while i < 101:
    print(i)
    i += 1

In the for loop the variable i is managed automatically. In the while loop we must create i ourselves, initialize it, and increment it inside the body.

Using a for loop:

for i in range(0, 100, 3):
    print(i)

Using a while loop:

i = 0
while i < 100:
    print(i)
    i += 3

Task: Output the sequence 3 5 7 9 … 21 twice — with a while loop and with a for loop.

Using a for loop:

for i in range(3, 21 + 1, 2):
    print(i, end=" ")

Using a while loop:

i = 3
while i <= 21:
    print(i, end=" ")
    i += 2

Output:

3 5 7 9 11 13 15 17 19 21

Task: Calculate the sum and the product of all odd two-digit integers: 11 + 13 + 15 + … + 99 and 11 × 13 × 15 × … × 99.

sum_ = 0
pr = 1
i = 11
while i <= 99:
    sum_ += i
    pr *= i
    i += 2
 
print(f"sum = {sum_} product = {pr}")

Expected output:

sum = 2565
product = 26480481588727626724487930522894928519708555015404150532676018390356298828125

Task: Read numbers and print their squares until -1 is entered.

num = int(input())
 
while num != -1:
    print(num * num)
    num = int(input())

Example run:

Input: 5    Output: 25
Input: 3    Output: 9
Input: -1   (Program ends)

Key takeaways:

• Initialization: the first number is read before the loop — it gives us the first value to check.
• Update: the next number is read at the end of the loop body. If we forgot this line, num would never change, and the loop would run forever!

Task: Sum numbers until the user types "stop".

text = input()
total = 0
 
while text != "stop":
    num = int(text)
    total += num
    text = input()
 
print("Sum of numbers is", total)

Example run:

2
5
3
stop
Sum of numbers is 10

Task: A sequence of words is provided on separate lines. The sequence ends when one of the words stop, enough, or quit appears. Those ending words are not counted. Output the total number of words in the sequence.

count = 0
word = input()
 
while word != "stop" and word != "enough" and word != "quit":
    count += 1
    word = input()
 
print(count)

Example run:

Skyrim
GTA
Mafia
stop
Battlefield
3

Task: Read integers divisible by 7, each on a separate line. The sequence ends as soon as a number that is not divisible by 7 appears. That terminating number is not part of the sequence — it only marks the end. Print all members of the sequence.

num = int(input())
 
while num % 7 == 0:
    print(num)
    num = int(input())

Example run:

49
2401
4809
0
2
10
100
Output:
49
2401
4809
0

The number 2 is not divisible by 7, so it stops the sequence and is not printed. Numbers after it are ignored.

i = 0
while i < 10:
    print("Hello")   # Missing i += 1, i is never updated!

The program prints "Hello" forever because i is never changed, and the condition i < 10 remains true.

Key takeaways:

• Infinite loops freeze your program.
• Always update the variables used in the condition inside the loop.

Interesting fact: In 1996 the Ariane 5 rocket exploded due to a software bug caused by a data overflow — a striking reminder of how dangerous an uncontrolled loop or overflow can be.

count = 0
 
while True:
    print(count)
    count += 1
    if count >= 10:
        break

Output:

0
1
2
3
4
5
6
7
8
9

The body executes at least once before the exit condition is checked. The break statement is covered in more depth in later lessons.

Task: Read numbers until 0 is entered and print the maximum.

maximum = -float('inf')
 
while True:
    num = int(input("Enter a number (0 to exit): "))
    if num == 0:
        break
    if num > maximum:
        maximum = num
 
print("Maximum:", maximum)

name = input()
 
while name != "Ivan" and name != "Arthur":
    print("Access Denied")
    name = input()
 
print("Access Granted")

Example run:

Maria
Access Denied
Olga
Access Denied
Oleg
Access Denied
Arthur
Access Granted

Task: Positive integers N and K are given. Using only addition and subtraction, find the quotient and the remainder of dividing N by K.

n = int(input('n? '))
k = int(input('k? '))
quot = 0
 
while k <= n:
    n -= k
    quot += 1
 
print()
print(f"quotient = {quot} remainder = {n}")

Example run:

n? 15
k? 7
quotient = 2 remainder = 1

Task: A pupil has n grades from 2 to 5. Calculate how many grades of each value were received.

n = int(input("Enter the number of grades: "))
count_2, count_3, count_4, count_5 = 0, 0, 0, 0
i = 0
 
while i < n:
    grade = int(input(f"Enter grade {i + 1}: "))
    if grade == 2:
        count_2 += 1
    elif grade == 3:
        count_3 += 1
    elif grade == 4:
        count_4 += 1
    elif grade == 5:
        count_5 += 1
    i += 1
 
print("Number of grades:")
print(f"2 = {count_2}")
print(f"3 = {count_3}")
print(f"4 = {count_4}")
print(f"5 = {count_5}")

Example run:

Enter the number of grades: 5
Enter grade 1: 2
Enter grade 2: 3
Enter grade 3: 4
Enter grade 4: 2
Enter grade 5: 3
Number of grades:
2 = 2
3 = 2
4 = 1
5 = 0

n = int(input())
 
while n != 0:
    digit = n % 10
    print(digit, end=' ')
    n = n // 10

Example run:

Input: 12345
Output: 5 4 3 2 1

Key takeaways: the standard pattern is extract, process, remove. The loop condition is n > 0 (or n != 0).

Task: Read a natural number and print its digits in a column in reverse order.

n = int(input())
 
while n != 0:
    print(n % 10)
    n = n // 10

Example run:

Input: 12345
Output:
5
4
3
2
1

Step-by-step explanation:

1. Read the integer n.
2. Set up a while loop that runs as long as n is not zero.
3. Inside the loop, n % 10 extracts the current last digit.
4. print() displays it on a new line.
5. n //= 10 removes the last digit, bringing the next digit into the "last" position.
6. The loop repeats until n becomes 0.

n = 123
product = 1
 
while n != 0:
    product *= n % 10
    n //= 10
 
print("Product of digits:", product)   # 6

Key takeaways: the sum uses addition (+=), the product uses multiplication (*=).

Task: Given a natural number, compute:

• the sum of its digits,
• the number of digits,
• the product of its digits,
• the average of its digits (as a float),
• its first digit,
• the sum of its first and last digit.

n = int(input())
original = n    # keep the original for first/last calculation
 
total = 0
count = 0
product = 1
 
while n != 0:
    digit = n % 10
    total += digit
    count += 1
    product *= digit
    n //= 10
 
# After the loop n is 0. Use the original to find the first digit.
first_digit = original // (10 ** (count - 1))
last_digit = original % 10
average = total / count
 
print("Sum:", total)
print("Count:", count)
print("Product:", product)
print("Average:", average)
print("First digit:", first_digit)
print("Sum of first and last:", first_digit + last_digit)

Example run (input: 5678):

Sum: 26
Count: 4
Product: 1680
Average: 6.5
First digit: 5
Sum of first and last: 13

Step-by-step explanation:

1. Initialize total = 0, count = 0, product = 1. Store original = n.
2. Loop: extract digits, add to total, increment count, multiply into product, remove the digit.
3. Post-processing:
   • first_digit = original // (10 ** (count - 1)). Since count is the number of digits, 10 ** (count - 1) gives the highest place value (e.g., 5678 // 1000 = 5).
   • last_digit = original % 10.
   • Average is total / count.

n = int(input("Input: "))
reversed_n = 0
 
while n != 0:
    digit = n % 10
    reversed_n = reversed_n * 10 + digit
    n //= 10
 
print("Output:", reversed_n)

Example run:

Input: 123
Output: 321

Trace for n = 123:

• Iteration 1: digit = 3, reversed_n = 0 × 10 + 3 = 3, n = 12
• Iteration 2: digit = 2, reversed_n = 3 × 10 + 2 = 32, n = 1
• Iteration 3: digit = 1, reversed_n = 32 × 10 + 1 = 321, n = 0

Task: Find the maximum and minimum digit of a number.

n = int(input())
 
max_digit = n % 10
min_digit = n % 10
n = n // 10
 
while n != 0:
    digit = n % 10
    if digit > max_digit:
        max_digit = digit
    if digit < min_digit:
        min_digit = digit
    n = n // 10
 
print("Max digit:", max_digit)
print("Min digit:", min_digit)

Example run (input: 53258):

Max digit: 8
Min digit: 2

Important note about zero: if n = 0 initially, the loop will not run at all. If you need to process 0 as a number with one digit, handle the case separately or use while True with a break condition.

Task: Determine whether a number consists of identical digits (e.g., 111 is YES, 112 is NO).

n = int(input())
 
if n == 0:
    print("YES")
else:
    last_digit = n % 10
    n //= 10
    is_identical = True
 
    while n != 0:
        digit = n % 10
        if digit != last_digit:
            is_identical = False
            break
        n //= 10
 
    if is_identical:
        print("YES")
    else:
        print("NO")

Example runs:

Input: 111111   Output: YES
Input: 111112   Output: NO

Step-by-step explanation:

1. Handle the n == 0 edge case.
2. Store the last digit as a reference (last_digit). Remove it from n.
3. Set a flag is_identical = True.
4. Loop through the remaining digits. If any digit differs from last_digit, set the flag to False and break out of the loop.
5. Print the result based on the flag.

num = 8619
n = len(str(num))
 
for i in range(1, n + 1):
    digit = num // 10 ** (n - i) % 10
    print(i, '-th digit is', digit)

Output:

1 -th digit is 8
2 -th digit is 6
3 -th digit is 1
4 -th digit is 9

Comparison with the while loop: the while loop extracts digits from right to left (last digit first), while this for loop extracts them from left to right. Choose the approach that fits your task.