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Find the largest k integer such that
$2^{k}$ divides $23^{2^{2016}}+7^{2^{2016}}$

@Ranler-Ranler
Suppose the existence of $k$ such that $2^{k}$ divides the RHS. WLOG suppose that $k \geq 2$ then 4 must divides the RHS which is clearly contradiction because RHS=2[4] then
$k \leq 2$ then $k = 1$ . ( $k = 2$ is Impossible).
Done!