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