Logic to check the number is divisible by 3 or not?

Question

without using %, / or * , I have to find the no. is divisible by 3 or not?

it might be an interview question.

Thanks.

Answer 1

There are various ways. The simplest is pretty obvious:

int isdivby3(int n) {
    if (n < 0) n = -n;

    while (n > 0) n -= 3;

    return n == 0;
}

But we can improve that. Any number can be represented like this: ("," means range inclusive):

Base2 (AKA binary)
(0,1) + 2*(0,1) + 4*(0,1)

Base4
(0,3) + 4*(0,3) + 16*(0,3)

BaseN
(0,N-1) + N*(0,N-1) + N*N*(0,N-1)

Now the trick is, a number x is divisible by n-1 if and only if the digitsum of x in base n is divisible by n-1. This trick is well-known for 9:

1926 = 6 + 2*10 + 9*100 + 1*1000
6+2+9+1 = 8 + 1*10
8+1 = 9 thus 1926 is divisible by 9

Now we can apply that for 3 too in base4. And were lucky since 4 is a power of 2 we can do binary bitwise operations. I use the notation number(base).

27(10) = 123(4)
Digitsum
12(4)
Digitsum again
3(4) = Divisible!

Now let's translate that to C:

int div3(int n) {
    if (n < 0) n = -n;
    else if (n == 0) return 1;

    while (n > 3) {
        int d = 0;

        while (n > 0) {
            d += n & 3;
            n >>= 2;
        }

        n = d;
    }

    return n == 3;
}

Blazing fast.