2013/14 AHSMC Part 1 Problem 10

Since $16^{2013} = 2^{4 \times 2013}$, $a$ must be of the form $2^k$ where $k$ is a positive integer divisor of $4 \times 2013 = 2^2 \times 3 \times 11 \times 61$. The prime factorization of $k$ may contain up to two $2$s, one $3$, one $11$ and one $61$, so that there are $(2+1)(1+1)^3 = 24$ possible values of $k$, and therefore of $a$. The answer is (E).