How to make a program for this in C?

Question

Take a number as input. print whether it is divisible by 2,3,5. use only one printf.

Answer 1

https://code.sololearn.com/c43R0zCswwJP/?ref=app

Answer 2

I wish it were Python. It'd have been easy

Answer 3

Use an if then add a printf

Answer 4

printf("%d is divisible by %s\n",n % 2 ? (n % 3 ? (n % 5 ? "neither 5, 3 nor 2" : "5") : "3") : "2"));

Answer 5

@Prunier 😂😂

Answer 6

Int ar[3]={2,3,5};
Int num; //number
For(i=0;i<=2;i++)
{
If(num%a[i]==0)
Printf("num is divisible by %d", a[i]) ;
Else
Continue;
}

Answer 7

well baptsie,ur program has same thing. if I put 30 it just returns 2.

Answer 8

kartikey can u type it nd give
I didn't understood u properly

Answer 9

I updated the program, see that

Answer 10

It is for each loop, the loop will iterate through each element in array.

Answer 11

continue skips the next part and jumps to run the next iteration

Answer 12

😊

Answer 13

#include <string.h>

...

char s[100];

printf("%d is divisible by %s\n",n % 2 && n % 3 && n % 5 ? "neither 5, 3 nor 2" : strcat(strcat(strcat(s, n%2 ? "" : "2 "), n % 3 ? "" : "3 "), n % 5 ? "":"5 "));

But once again, do not do this at home :p

Answer 14

At least, only one printf will be called and written :p

Answer 15

@Baptiste this code wrks i think😐

Answer 16

Oki oki i gt to know

Answer 17

I tried. but it doesn't work if no.
like 30. it prints only one of 2,3,5.

Answer 18

how to do it?

Answer 19

/* let's say the number is n */
if(n % 2){
if(n % 3){
if(n % 5)
printf("%d is neither divisible by 2, 3 nor 5\n",n);
else
printf("%d is divisible by 5\n",n);
}else
printf("%d is divisible by 3\n",n);
}else
printf("%d is divisible by 2\n",n);

Other solution :

int divisible = 0;

if(!(n%5)){
printf("%d is divisible by 5\n",n);
divisible = 1;
}

if(!(n%3)){
printf("%d is divisible by 3\n",n);
divisible = 1;
}

if(!(n&1)){
printf("%d is divisible by 2\n",n);
divisible = 1;
}

if(!divisible)
printf("%d is neither divisible by 2, 3 nor 5\n",n);

Answer 20

I said only 1 printf