How many positive integers 'n' are there such that [(2)^n + 1] is divisible by 7 ?

Zero.

The remainder of the (2^n+1)/7 follows the pattern 2,3,5,2,3,5. This will never reach zero to create a solution. If you take one of the remainders, subtract 1, double it, and add 1, then the next number in the series appears. Thus there are zero solutions for this problem.

Posted by Dan Blume on 2004-01-02 16:44:11