C Program to Find LCM and GCD of Two Numbers | C Program

Here's a C program to find LCM and GCD of the given two numbers using while loop and IF condition with output and proper explanation.

# include <stdio.h>
# include <conio.h>
void main() 
{ 
 int n1, n2, prod, gcd, lcm ; 
 clrscr() ; 
 printf("Enter the two numbers : ") ; 
 scanf("%d %d", &n1, &n2) ; 
 prod = n1 * n2 ; 
 while(n1 != n2) 
 { 
  if(n1 > n2) 
   n1 = n1 - n2 ; 
  if(n2 > n1) 
   n2 = n2 - n1 ; 
 } 
 gcd = n1 ; 
 lcm = prod / gcd ; 
 printf("\nThe GCD is : %d", gcd) ; 
 printf("\n\nThe LCM is : %d", lcm); 
 getch() ; 
} 

Output of above program is

Enter the two numbers : 10       8
The GCD is : 2

The LCM is : 40

Explanation of above program

• GCD Calculation -

We're calculating GCD of two numbers n1 and n2 inside the while loop. The while loop wil run until both the numbers n1 and n2 become equal. The process of gcd calculation is as follows -

1. If n1 > n2 then update the value of n1 as - n1 = n1 - n2.
2. If n2 > n1 then update the value of n2 as - n2 = n2 - n1.
3. If n1 != n2 then repeat step 1 and 2 otherwise n1 or n2 is the value of GCD.

• LCM Calculation - 

LCM can be calculated using a simple formula -

LCM = Product of numbers /  GCD of numbers = n1 * n2 / gcd.

Suppose, n1 = 10 and n2 = 8 then steps of GCD calculation are as follows -

• n1 > n2 i.e. 10 > 8 so n1 = n1 - n2 = 10 - 8 = 2. Therefore now, n1 = 2 and n2 = 8.
• n2 > n1 i.e. 8 > 2 so n2 = n2 - n1 = 8 - 2 = 6. Therefore now, n1 = 2 and n2 = 6.
• n2 > n1 i.e 6 > 2 so n2 = n2 - n1 = 6 - 2 = 4. Therefore now, n1 = 2 and n2 = 4.
• n2 > n1 i.e. 4 > 2 so n2 = n2 - n1 = 4 - 2 = 2. Therefore now, n1 = 2 and n2 = 2.
• Now n1 = n2 so gcd = n1 or n2 = 2.

Also LCM = Product / gcd = 10 * 8 / 2 = 40.