Describe all n such that 3 devides n2^n+1

[rbetan]

Describe All n such that 3 divides n times 2 raised to the n, plus one.( n*2^n + 1)

I know that a number is divisible by 3 if the sum of its digits adds up to a number that itself is divisible by 3. But this is probably not helpful for this problem.

I also know that a number A is divisible by B if when A is divided by B the remainder is 0. In other words, A is congruent to 0 in mod B.

My original idea is to assume that n*2^n + 1= 0(mod 3). And then somehow try to see for which values of n this is possible. But i am stuck.

Maybe this is not the right approach.

Any help with this will be highly appreciated. Thanks

[ramsey2879]

Break the problem down to smaller steps what is the sequence of 2^n mod 3? Then what is n time this plus 1? P.S. The period of the sequence for n mod 3 is different from the period for 2^n mod 3. Their product is the period for the description of n*2^n + 1. With this in mind your original approach is correct.