### 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:

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