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

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

#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;
}

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

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