Check if a number is divisible by 3 [closed]

Answer 1

Here is an simple way to do it by hand. Since 1 = 2² mod 3, we get 1 = 2²ⁿ mod 3 for every positive integer. Furthermore 2 = 2²ⁿ⁺¹ mod 3. Hence one can determine if an integer is divisible by 3 by counting the 1 bits at odd bit positions, multiply this number by 2, add the number of 1-bits at even bit posistions add them to the result and check if the result is divisible by 3.

Example: 57₁₀=111001₂. There are 2 bits at odd positions, and 2 bits at even positions. 2*2 + 2 = 6 is divisible by 3. Hence 57 is divisible by 3.

Here is also a thought towards solving question c). If one inverts the bit order of a binary integer then all the bits remain at even/odd positions or all bits change. Hence inverting the order of the bits of an integer n results is an integer that is divisible by 3 if and only if n is divisible by 3. Hence any solution for question a) works without changes for question b) and vice versa. Hmm, maybe this could help to figure out which approach is faster...