Answer / sriharsha

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 / k.kavitha

main()
{
int num=0,sum=0,k=0;
pirntf("enter the number\n");
scanf("%d",num);
while(num!=0);
{
k=num%10;
sum=sum+k;
num=num/10;
}
printf("%d",sum);
}

Answer / rashmi sharma

#include<sthio.h>
#include<conio.h>
void main()
{
int a,b,c,d,e,f,g,h,i,j,k,l;
printf("enter a five digit number");
scanf("%d",&a);
b=a%10;
c=a/10;
d=c%10;
e=c/10;
f=e%10;
g=e/10;
h=g%10;
i=g/10;
j=i%10;
k=i/10;
l=b+d+f+h+j;
printf("sum of the digit=%d",l);
getch();
}

Answer / joy

#include<stdio.h>

main ()
{
int num, a, b,c,d,e, sum;

printf ("Enter the five digit number : ");
scanf ("%d", &num);
e = num % 10;
d = (num / 10)% 10;
c = (num / 100)% 10;
b = (num / 1000)% 10;
a = (num / 10000)% 10;
sum= a+b+c+d+e;
printf ("The sum of the digits is: %d", sum);

}

Answer / jyoti

#include<stdio.h>
int main()
{
int n,d1,d2,d3,d4,d5,s;
printf("enter a five digit no");
scanf("%d",&n);
d5=n%10;n=n/10;
d4=n%10;n=n/10;
d3=n%10;n=n/10;
d2=n%10;n=n/10;
d1=n%10;
s=d1+d2+d3+d4+d5;
return 0;
}

Answer / md osama

#include<stdio.h>
#include<conio.h>
void main()
{
int d1,d2,d3,d4,d5,sum;
long num;
clrscr();
printf("enter the five digit number:");
scanf("%ld",&num);
d1=(num%10);
d2=(num%100-num%10)/10;
d3=(num%1000-num%100)/100;
d4=(num%10000-num%1000)/1000;
d5=(num-num%10000)/10000;
sum=d1+d2+d3+d4+d5;
printf("\n the sum of the digit is:%d",sum);
}

Answer / richa

main()
{
int num , sum=0 ;
printf(" enter 5 digit no. " );
scanf(" %d " , num);
while (num!=0)
{ sum+=num%10;
num=num/10;
}
printf("sum is = %d " , sum );
getch();
}

Answer / haris

Try it. It will seriously work. You would appreciate it.

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

int main()

{
int x;
int a;
int b;
int c;
int d;
int e;
int sum;

printf("Please enter a five digit number:\n");
scanf("%d",&x);

a=x%10;
b=(x/10)% 10;
c=(x/100)% 10;
d=(x/1000)%10;
e=(x/10000)%10;

sum=a+b+c+d+e;

printf("\n\nThe sum of all digits is %d.",sum);

getch();
return 0;

}

Answer / jegadeesh

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

}

//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 answer u are looking...

Answer / areeb ahmed khan

#include<stdio.h>
#include<conio.h>
main()
{
int a,b,c,d,e;
int digit,number,sum;

printf("\n\n\n");

printf("Enter a five-digit number :");
scanf("%d",&number);

digit=number%10;
a=digit;

number=number/10;
digit=number%10;
b=digit;

number=number/10;
digit=number%10;
c=digit;

number=number/10;
digit=number%10;
d=digit;

number=number/10;
digit=number%10;
e=digit;

sum=a+b+c+d+e;
printf("\nThe sum of five-digit number is : %d",sum);

getch();
}

What are different types of operators?

0 Answers  

main() { int a,b; printf("%d,%d",scanf("%d%d",&a,&b)); } => do u mean above program's output... =>output will be:2,whatever you enter value for b. =>because scanf is a library fn which will return how many arguements it processes, and second value you are right mr.Satya but i found my self unable to understand that for the first time scanf returns the no of successful matches but how for the second time it returns the value of 'b'.while a function should return the same 'r' value every time.

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PLS U SENS ME INTERVIEW O. MY EMAIL ADD, SOFIYA.SINGH@GMAIL.COM

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#include<stdio.h> #include<conio.h> void main() { clrscr(); int a=0,b=0,c=0; printf("enter value of a,b"); scanf(" %d %d",a,b); c=a+b; printf("sum is %d",c); getch(); }

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