Hmm, I wonder if this is more efficient than calculating mod 15 directly. It's basically something like gcd15(int x) { return ((x * 0xaaaaaaab) <

Hmm, I wonder if this is more efficient than calculating mod 15 directly. It's basically something like gcd15(int x) { return ((x * 0xaaaaaaab) < (0xfffffff/3 + 1) ? 1 : 3) * ((x * 0xaaaaaaab) < (0xfffffff/5 + 1) ? 1 : 5) } versus something like gcd15(int x) { LOOKUP_TABLE = { /* gcd(n,15) for n = 0,...,15 */ } int q = (x * ((1 << 28) / 15)) >> 28; return LOOKUP_TABLE[x - 15 * q]; } Well, you're probably better of using the method in the article for the division, but this should work for small numbers.