Program 33:To Print GCD and LCM

Program 33:
 

#include<stdio.h>
main()
{
int num1,num2,temp1,temp2,gcd,lcm,x;
printf("Enter number 1\n");
scanf("%d",&num1);
printf("Enter number 2\n");
scanf("%d",&num2);

temp1=num1;
temp2=num2;

while(temp2!=0)
{
x=temp2;
temp2=temp1%temp2;
temp1=x;
}

gcd=temp1;
lcm=(num1*num2)/gcd;
printf("gcd is %d\n",gcd);
printf("Lcm is %d\n",lcm);

}

Explanation:

1. The program starts with initializing
   • gcd → For storing the gcd result
   • num1,num2 → For storing two integers
   • lcm→ For storing the lcm.
   • temp1,temp2 → For storing num1 and num2 values temporarily

2. printf("Enter number 1\n");
   scanf("%d",&num1);
   printf("Enter number 2\n");
   scanf("%d",&num2);
   

   For taking 2 integers to find gcd and lcm

3. temp1=num1;
   temp2=num2;
   

   now temp variables sore values of 2 integers

4. while(temp2!=0)
   {
   x=temp2;
   temp2=temp1%temp2;
   temp1=x;
   }

   • Let us assume num1=11,num2=13,then temp1=11,temp2=13
   • Let us start iteration:
     • step1: as temp2 is not zero → true so loop continues iteration
       • x=temp2→ x=13
       • temp2=11%13 →temp2=11
       • temp1=13
       • The final values are temp1=13,temp2=11
     • step2: as temp2!=0 → true so loop continues iteration
       • x=temp2→ x=11
       • temp2=13%11 →temp2=2
       • temp1=11
       • The final values are temp1=11,temp2=2
     • step3: as temp2!=0 → true so loop continues iteration
       • x=temp2→ x=2
       • temp2=11%2 →temp2=1
       • temp1=2
       • The final values are temp1=2,temp2=1
     • step4: as temp2!=0 → true so loop continues iteration
       • x=temp2→ x=1
       • temp2=2%1 →temp2=0
       • temp1=1
       • The final values are temp1=1,temp2=0
     • step4: as temp2!=0 → false so loop terminates

5. gcd=temp1;
   lcm=(num1*num2)/gcd;
   printf("gcd is %d\n",gcd);
   printf("Lcm is %d\n",lcm);

   gcd=temp1 → gcd=1 from above. lcm=(11*13)/1 → lcm=143.
6. Then the loop prints gcd and lcm.

Output: