Basic if else question

I have to create a program that takes 2 numbers. If the first number is equally divisible by the second number, then print the number of times it's divisible, otherwise, print a message saying "not divisible". I have no clue how to write an if statement to find out if the first number is equally divisible by the second number.

My code is below - please explain.

Code:

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#include <stdio.h>

int main()
{
    int x, y, z;
   
   
    printf("Enter in 2 numbers: \n");
    scanf("%d%d", &x, &y);
   
    if
      {x % y ==z}
      {
            printf("%d\n", z);
      }
            else
            {
                printf("not divisible\n");
            }
    }
   
    return 0;
}

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Quote:

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Originally Posted by benrogers

Code:

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#include <stdio.h>

int main()
{
    int x, y, z;
   
   
    printf("Enter in 2 numbers: \n");
    scanf("%d%d", &x, &y);
   
    if
      {x % y ==z}
      {
            printf("%d\n", z);
      }
            else
            {
                printf("not divisible\n");
            }
    }
   
    return 0;
}

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bold does the trick.

Code:

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#include <stdio.h>

int main()
{
    int x, y, z;
   
    printf("Enter in 2 numbers: \n");
    scanf("%d%d", &x, &y);
 
    z = x / y;
 
    if (x % y == 0)
    {
          printf("%d\n",z);
          }
    else
    {
        printf("not divisible\n");
    }
   
    return 0;
}

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Code:

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while (!(num1 % num2))
{
    count++ ;
      num1/=2 ;
}

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i think that would work

1. You didn't assign z to anything.
2. The expression in the if statement should be in parenthesis, not braces.
3. For future reference, you should write code that catches division by zero, like this.

Code:

---

#include <stdio.h>

int main()
{
    int x, y, z;
   
    printf("Enter in 2 numbers: \n");
    scanf("%d%d", &x, &y);
 
    if (y != 0) z = x / y;
    else {
        printf("division by zero\n");
        return 1;
    }
    if (x % y == 0)
    {
          printf("%d\n",z);
          }
    else
    {
        printf("not divisible\n");
    }
   
    return 0;
}

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