Answer / vani

//sum of digits in c//
#include<stdio.h>
#include<conio.h>
void main()
{
int n,r,sum=0;
clrscr();
printf("\n enter the number:");
scanf("%d",&n);
while(n!=0)
{
r=n%10;
sum=sum+r;
n=n/10;
}
printf("sum=%d");
getch();
}

Answer / jay soni

#include<stdio.h>
#include<conio.h>
void main()

{
int n,sum=0,r;
printf("enter the value");
scanf("%d",&n);
while(n>0)
{
r=n%10;
n=n/10;
sum=sum+r;
}
printf("\nsum of digits is %d",sum);

getch();
}

Answer / giri

#include <stdio.h>
#include <conio.h>

void main()
{
int sum=0, num, mul=1;
printf("Enter the number");
scanf("%d", &num);

while(num)
{
sum = sum + ((num%10)*mul);
num = num/10;
mul = mul * 10;
}

printf("Sum is %d", sum);
}

Answer / rajan praksh more (vangani-tha

#include<stdio.h>
#include<conio.h>
void main()
{
int rem,a;
int sum=0;
clrscr();
printf("Enter The Number: ");
scanf("%d",&a);
while(a>0)
{
rem=a%10;
sum=sum+rem;
a=a/10;
}
printf("sum of digits of entered number is =%d",sum);
getch();
}

Answer / mokhtar

all your answers are wrong, guys, for example when you get the reminder of 2/10, it gives you 0, because all of you used int as a data type.
so I think the the method is right, but the implementation is wrong.

regards.

Answer / rakesh ranjan

#include<conio.h>
#include<stdio.h>
main()
{
int x=0,n,i;
printf("entre the number");
scanf("%d",&n);
for(;n>0;n/=10)
x=x+n%10;
printf("sum = %d",x);
getch();
}

Answer / gg

try this....

#include<stdio.h>
int main()
{
int n,m,sum=0;
scanf("%d",&n);
for(m=0;((m=n%10)!=0);n=(n/10))
sum+=m;
printf("count is %d\n",sum);
}

Answer / ruchi

#include<stdio.h>
#include<conio.h>
int main()
{
int n,m,d,m1,s=0,s1;
printf("enter the number ");
scanf("%d",&n);
m=n%10;
d=n/10;
while(d>=10)
{
m1=d%10;
d=d/10;
s=s+m1;
}
s1=s+m+d;
printf("\nThe sum of digits is ");
printf("%d",s1);
getch();
}

Answer / deepak upadhyay

#include<stdio.h>
void main()
{
int num,a,b,c,d,e,sum;
printf("enter the num");
scanf("%d",&num);
a=num%10;
b=((num%100)-a)/10;
c=((num%1000)-(num%100))/100;
d=((num%10000)-(num%1000))/1000;
e=((num%100000)-(num%10000))/10000;
sum=a+b+c+d+e;
printf("\n1's place= %d \n10's place= %d \n100's
place= %d \n1000's place= %d \n10000's place=
%d",a,b,c,d,e);
printf("\nsum=%d",sum);
}

Answer / leelanarasimhareddy

void main()
{
int a[20];
printf("enter no of values");
scanf("%d",&n);
for(i=0;i<n;i++)
{
scanf("%d",&a[i]);
}
for(i=0;i<n;i++)
{
sum=sum+a[i];
}
printf("%d",sum);
}

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