Divide a number by 3 without using *, /, +, -, % operators

Question

This is a simple function which performs the desired operation. But it requires the + operator, so all you have left to do is to add the values with bit-operators:

// replaces the + operator
int add(int x, int y)
{
    while (x) {
        int t = (x & y) << 1;
        y ^= x;
        x = t;
    }
    return y;
}

int divideby3(int num)
{
    int sum = 0;
    while (num > 3) {
        sum = add(num >> 2, sum);
        num = add(num >> 2, num & 3);
    }
    if (num == 3)
        sum = add(sum, 1);
    return sum; 
}

As Jim commented this works, because:

• n = 4 * a + b
• n / 3 = a + (a + b) / 3

• So sum += a, n = a + b, and iterate

• When a == 0 (n < 4), sum += floor(n / 3); i.e. 1, if n == 3, else 0

Idiotic conditions call for an idiotic solution:

#include <stdio.h>
#include <stdlib.h>

int main()
{
    FILE * fp=fopen("temp.dat","w+b");
    int number=12346;
    int divisor=3;
    char * buf = calloc(number,1);
    fwrite(buf,number,1,fp);
    rewind(fp);
    int result=fread(buf,divisor,number,fp);
    printf("%d / %d = %d", number, divisor, result);
    free(buf);
    fclose(fp);
    return 0;
}

If also the decimal part is needed, just declare result as double and add to it the result of fmod(number,divisor).

Explanation of how it works

1. The fwrite writes number bytes (number being 123456 in the example above).
2. rewind resets the file pointer to the front of the file.
3. fread reads a maximum of number "records" that are divisor in length from the file, and returns the number of elements it read.

If you write 30 bytes then read back the file in units of 3, you get 10 "units". 30 / 3 = 10

log(pow(exp(number),0.33333333333333333333)) /* :-) */

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[])
{

    int num = 1234567;
    int den = 3;
    div_t r = div(num,den); // div() is a standard C function.
    printf("%d\n", r.quot);

    return 0;
}

You can use (platform dependent) inline assembly, e.g., for x86: (also works for negative numbers)

#include <stdio.h>

int main() {
  int dividend = -42, divisor = 5, quotient, remainder;

  __asm__ ( "cdq; idivl %%ebx;"
          : "=a" (quotient), "=d" (remainder)
          : "a"  (dividend), "b"  (divisor)
          : );

  printf("%i / %i = %i, remainder: %i\n", dividend, divisor, quotient, remainder);
  return 0;
}

Use itoa to convert to a base 3 string. Drop the last trit and convert back to base 10.

// Note: itoa is non-standard but actual implementations
// don't seem to handle negative when base != 10.
int div3(int i) {
    char str[42];
    sprintf(str, "%d", INT_MIN); // Put minus sign at str[0]
    if (i>0)                     // Remove sign if positive
        str[0] = ' ';
    itoa(abs(i), &str[1], 3);    // Put ternary absolute value starting at str[1]
    str[strlen(&str[1])] = '\0'; // Drop last digit
    return strtol(str, NULL, 3); // Read back result
}