PROCEDURE:-
1.enter integer value n
2.repeat step 3 to 8 until n is greater than 9
3.set sum to 0
4.repeat step 5 to 7 until n is greater than
5. find the modulo 10 for n and store result in r
6.add r to previous sum
7. find new value of n by dividing it by 10
8 set n to sum
9. print sum
CODE:-
#include<stdio.h>
void main()
{
int n,m,sum,r;
printf(“enter value for n”);
scanf(“%d”,&n);
m=n;
while(n>9)
{
sum=0;
while(n>0)
{
r=n%10;
sum=sum+r;
n=n/10;
}
n=sum;
}
printf(“sum of %d down to single digit is %d”,m,sum);
}
Input:- enter value for n 137
Output:- sum of 137 down to single digit is 2
1.enter integer value n
2.repeat step 3 to 8 until n is greater than 9
3.set sum to 0
4.repeat step 5 to 7 until n is greater than
5. find the modulo 10 for n and store result in r
6.add r to previous sum
7. find new value of n by dividing it by 10
8 set n to sum
9. print sum
CODE:-
#include<stdio.h>
void main()
{
int n,m,sum,r;
printf(“enter value for n”);
scanf(“%d”,&n);
m=n;
while(n>9)
{
sum=0;
while(n>0)
{
r=n%10;
sum=sum+r;
n=n/10;
}
n=sum;
}
printf(“sum of %d down to single digit is %d”,m,sum);
}
Input:- enter value for n 137
Output:- sum of 137 down to single digit is 2
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