# Program to add the numbers which are divisible by 5 in C

Let us write a program to print all the numbers between N1 and N2 which are divisible by 5. Then, compute the sum of those numbers. The numbers between N1 and N2 can be added using the following statement:

sum =0;
for(i=n1; i<=n2; i++)       /* i start from n1 and goes up to n2 */
{
sum =sum+i;                    /* Can be written as sum +=i */
}

But, it is required to add those which are divisible by 5 only. This can be achieved by taking only those values of i which are divisible by 5 and the program can be modified as follows:

sum =0;
for(i=n1; i<=n2; i++)
{
if(i%5 == 0)
{
sum +=i;
}
}

The complete algorithm are shown below:

Algorithm : DIVBY5.

[This algorithm prints the numbers divisible by 5 between n1 and n2]
Step 1: [Enter lower & upper limits]

Step 2: [Initialize]
sum =0

Step 3: [Generate and add the number if divisible by 5]
for i=n1 to n2 in step 1
if(i%5 ==0)
sum = sum+i
[End of if]
[End of for]

Step 4: [Output the result]
Write : sum

Step 5: [Finished]
Exit.

The complete program to find the sum of all those numbers divisible by 5 is shown below:
Example : Program to add the numbers which are divisible by 5.
PROGRAM

#include<stdio.h>
main( )
{
int n1, n2, i, sum;
printf("Enter range N1 and N2 (N1 < N2): \n");
scanf("%d%d",&n1,&n2);
sum=0;
for(i=n1;i<=n2;i++)
{
if((i%5)==0)
{
printf("%d",i);
sum +=i;
}
}
printf("Total sum =%d\n",sum);
}

### TRACING

Execution starts from here
Non-executable statement
Input
Enter the N1 and N2 (N1<N2):
4 20
Computations
sum=0
i= 5 10 15 20
Output
5 10 15 20
sum = 0+5+10+15+20

Total sum = 50

Some say he’s half man half fish, others say he’s more of a seventy/thirty split. Either way he’s a fishy bastard. Google