Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Jun 13, 2009 #1
    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
  2. jcsd
  3. Jun 13, 2009 #2
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook