Determine if an integer is divisible by 3

C - 2 tokens

int div3(int x) {
    return x * 0xAAAAAAAB <= x;
}

Seems to work up to 2³¹-1.

Credits to zalgo("nhahtdh") for the multiplicative inverse idea.

Python, 3 2 tokens

Brute force solution, but it works.

0x9249249249249249249249249249249249249249249249249249249249249249>>x&1

Thanks to Howard for the 1 token reduction.

C - 5 4 (?) tokens

int div3_m2(uint32_t n) {
    return n == 3 * (n * 0xAAAAAAABull >> 33);
}

Works for any unsigned 32-bit number.

This code makes use of multiplicative inverse modulo 2³² of a divisor to convert division operation into multiplication operation.

Edit

My solution (posted 2 minutes after) has the same spirit as aditsu's solution. Credit to him for the use of == that improves my solution by 1 token.

Reference

• Labor of Division - ridiculousfish.com
• Hacker's Delight - Magic Number
• Wikipedia - Modular multiplicative inverse