Answer / synner

#include<stdio.h>
#include<math.h>
#include<conio.h>
void main()
{
long int num;
int sum=0,temp;
clrscr();
printf("Enter the numbe\n");
scanf("%ld",&num);
while(num!=0)
{
temp=num%10;
sum=sum+temp;
num=num/10;
}
printf("%d",sum);
getch();
}

Answer / aha na

#include<stdio.h>
#include<conio.h>
void main()
{
int i,j,sum=0;
clrscr();
printf("\n enter a number",i);
scanf("%d",&i);

while(i<=0)
{
j=i%10;
sum=sum+j;
i=i/10;
}
printf("\n the sum of the digits
are:%d\n");
getch();
}

Answer / jegadeesh

#include<stdio.h>
void main()
{
int i,j;
scanf("%d",&i);
if((i%9)==0)
printf("The sum of digit is 9");
else
{
j=i%9;
printf("%d",j);
}
}

//the above will print result as 6 if given as 12345 since
1+2+3+4+5=15 and proceeding 1+5 we get as 6.....
is this an apt answer...

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