algorithm - How to know if a binary number divides by 3?

Question

I want to know is there any divisible rule in binary system for dividing by 3.

For example: in decimal, if the digits sum is divided by 3 then the number is devided by 3. For exmaple: 15 -> 1+5 = 6 -> 6 is divided by 3 so 15 is divided by 3.

The important thing to understand is that im not looking for a CODE that will do so.. bool flag = (i%3==0); is'nt the answer I'm looking for. I look for somthing which is easy for human to do just as the decimal law.

Answer 1

Refer to this website: How to Tell if a Binary Number is Divisible by Three

Basically count the number of non-zero odd positions bits and non-zero even position bits from the right. If their difference is divisible by 3, then the number is divisible by 3.

For example:

15 = 1111 which has 2 odd and 2 even non-zero bits. The difference is 0. Thus 15 is divisible by 3.

185 = 10111001 which has 2 odd non-zero bits and 3 even non-zero bits. The difference is 1. Thus 185 is not divisible by 3.

Explanation

Consider the 2^n values. We know that 2^0 = 1 is congruent 1 mod 3. Thus 2^1 = 2 is congurent 2*1 = 2 mod 3. Continuing the pattern, we notice that for 2^n where n is odd, 2^n is congruent 1 mod 3 and for even it is congruent 2 mod 3 which is -1 mod 3. Thus 10111001 is congruent 1*1 + 0*-1 + 1*1 + 1*-1 + 1*1 + 0*-1 + 0*1 + 1*-1 mod 3 which is congruent 1 mod 3. Thus 185 is not divisible by 3.