Answer / neo

#include <stdio.h>

void div(int n){
int i=2;
while(n%i!=0 && i!=n){
i++;
}
printf("%d ",i);
if(i!=n){
div(n/i);
}

}

main(){
int i;
printf("Enter number:");scanf("%d",&i);
printf("1 ");
div(i);
}

Answer / rakesh ranjan

#include<conio.h>
#include<stdio.h>
main()
{
int n,i;
printf("entre the number");
scanf("%d",&n);
for(i=2;i<n;i++)
{
if(n%i==0)
printf("%d\n",i);
}
getch();
}

Answer / dally

#include<stdio.h>
int main()
{
int n,i=1;
printf("Value for n\n");
scanf("%d\n",&n);
while(i<=n)
{
if(n%i == 0)
printf("%d\n",i);

i++;
}
}

Answer / tamil

void div(int n){
static int i=1;
while(i<=n){
if(!(n%i))printf(" %d",i);
i++;
}
}

main(){
int i;
clrscr();
printf("Enter number:");scanf("%d",&i);
div(i);
getch();
}

Answer / guest

Optimised!! :-) some extra condition added to avoid printing repeated numbers.

#include<stdio.h>
void dev(int n,int i)
{
if(n <= i) return;
while(i <= n){
if((n % i) == 0){
if(n!=i) printf("%d ",i);
printf("%d ",n/i);
break;
}
i++;
}
dev(n/i,i+1);
return;
}
main()
{
int n;

printf("Enter number:");
scanf("%d",&n);

dev(n,2);
printf("\n");
return 0;
}

Answer / pradeep

same program as above, optimising it a little bit.

#include<stdio.h>
int main()
{
int n,i=1;
printf("Value for n\n");
scanf("%d\n",&n);
while(i <= n/2)
{
if(n%i == 0)
printf("%d\n",i);

i++;
}
printf("%d\n",n);
}

Answer / buya

111,111,111*111,111,111=12345678910987654321

write a programe to accept any two number and check the following condition using goto state ment.if a>b,print a & find whether it is even or odd and then print.and a<b,printb.find the sum digits of that number & then print.if a==b multiply 10 with a & add 20 with b store in c and then print

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