# Binary 2^n - 1 divisibility

• kurtulmehtap
In summary, the answer to whether a binary number is divisible by three is either testing if the number is a multiple of 11 or 10, or subtracting then adding the bits together.

#### kurtulmehtap

Dear All,
I have to test if a binary number is divisible to
2^n - 1 where n is even.
Is there a test available for binary numbers like to test a divisibility by 3.

You are talking about what, in general, is studied as Mersenne primes. Even powers are generally not prime. That's not hard to show.

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The sum of the base 2^n digits is divisible by 2^n-1 iff the number is, so you could use that. But I'm not sure how much of a speedup that gives.

robert Ihnot said:
You are talking about what, in general, is studied as Mersenne primes. Even powers are generally not prime. That's not hard to show.

I interpreted the question as "how can I tell if N is divisible by 2^n-1" rather than "how can I tell if N divides 2^n-1".

robert Ihnot said:
You are talking about what, in general, is studied as Mersenne primes. Even powers are generally not prime. That's not hard to show.

in my case n is even, not odd or prime

Testing if a binary number is divisible by three is like testing if a number base 10 is divisible by 11. Alternately subtract than add the bits together and if your answer is divisable by three than the whole string was divisable by three.

1100101 1-1+0-0+1-0+1=2 not divisible by 3
1100110 1-1+0-0+1-1+0=0 divisible by three.

robert2734 said:
Testing if a binary number is divisible by three is like testing if a number base 10 is divisible by 11. Alternately subtract than add the bits together and if your answer is divisable by three than the whole string was divisable by three.

1100101 1-1+0-0+1-0+1=2 not divisible by 3
1100110 1-1+0-0+1-1+0=0 divisible by three.

3 is a funny number, since it's 2^1 + 1 and 2^2 - 1. You can use either type of test: add digits in blocks of two, or alternately add and subtract digits.

10101011 = 10+10+10+11 = 1001 = 10 + 01 = 11 = 0 (mod 3)
10101011 -> 1-0+1-0+1-0+1-1 = 11 = 0 (mod 3)

The first preserves residues (not just divisibility) mod 3; the second works with smaller numbers (though twice as many).

You can test if a number base 4 is a multiple of three the way you test if a number base ten is a multiple of nine. I saw your earlier post and decided to add some additional information. I don't know which approach , if either, gives you any calculating speed up.