Think of bit-operators that can help you carry out operations.

There are a few solutions for this, the most accepted answer though has been copied from a StackOverflow question below:

int add(int x, int y) {
    int a, b;
    do {
        a = x & y;
        b = x ^ y;
        x = a << 1;
        y = b;
    } while (a);
    return b;
}

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; 
}

This works because:
1. n = 4 * a + b
2. n / 3 = a + (a + b) / 3
3. So sum += a, n = a + b, and iterate.
4. When a == 0 (n < 4), sum += floor(n / 3); i.e. 1, if n == 3, else 0

Again please view the original question and answer from StackOverflow.

Thoughts or alternate solution? Let us know in the comments below!