Python program to find h.c.f. of two numbers using iteration

In this tutorial, we are going to learn writing a python program to calculate the Highest Common Factor of two numbers. In this program we will use iteration concept.

Problem Statement

For any two numbers that are inputs given by the user, we have to calculate and print the h.c.f. of that numbers.

 For example:

Case1: If the user inputs the numbers 4 and 6.

            then the output should be ‘2’.

Case2: If the user inputs the numbers 5 and 7.

            then the output should be ‘1’.

Our logic to find h.c.f. of two numbers

• Our program will take two inputs numbers from the user.
• Using the ‘if-else’ conditional statement, we find the minimum of two input integers and store it in the ‘minimum’ variable.
• We will use ‘for loop’ with ‘range’ function with the argument, 1 as start, and minimum+1 as the stop.
• Inside the ‘for loop’ block, we find the highest factor which is common to both input numbers.
• Print the HCF as the output of our program.

Python code to find h.c.f. of two numbers

#taking two inputs from the user
num1 = int(input("Enter first number: "))  
num2 = int(input("Enter second number: "))  
#checking for smaller number
if num1 > num2:
    minimum = num2
else:
    minimum = num1
#finding highest factor of the numbers
for i in range(1, minimum+1):
    if((num1 % i == 0) and (num2 % i == 0)):
        hcf = i 
print("hcf/gcd of",num1,"and",num2,"=",hcf)

Output 1:

Enter first number: 8
Enter second number: 2
hcf/gcd of 8 and 2 = 2

Explanation:

 For inputs 8 and 2, the program is finding the highest factor that is common in both 8 and 2.

The factor of 8 is 1,2,4,8 and the factor of 2 is 1,2. The highest factor common in both is 2.

Output 2:

Enter first number: 25
Enter second number: 18
hcf/gcd of 25 and 18 = 1

Explanation:

For inputs 25 and 18, the program is finding the highest factor that is common in both 25 and 18.

The factor of 25 is 1,5,25 and the factor of 18 is 1,2,3,6,9,18. The highest factor common in both is 1.