2015/16 AHSMC Part 1 Problem 11

We should have $5n+1 = \pm 2^k$ where $0 \le k \le 2015$. If $5n+1 = 2^k$, then $5 \mid (2^k-1)$, which happens if $k = 0,4,8,\ldots$, hence $k = 4s$, $0 \le s \le 503$. For these values of $k$ we obtain $504$ nonnegative integers $n$. If $5n+1 = -2^k$ then $5 \mid (2^k+1)$, which happens if $k = 2,6,10,\ldots$, hence $k = 4s+2$, $0 \le s \le 503$. For these values of $k$ we obtain $504$ negative integers $n$. Therefore there are $1008$ convenient values for $n$. The answer is (E).