Romanian Mathematical Olympiad

The number of positive divisors of an integer $N$ with the prime factorization $p_1^{a_1} p_2^{a_2} \cdots p_n^{a_n}$ is $(a_1 + 1)(a_2 + 1) \cdots (a_n + 1)$. Since $2015 = 1 \cdot 2015 = 5 \cdot 403 = 13 \cdot 155 = 31 \cdot 65 = 5 \cdot 13 \cdot 31$, the required number has one of the following forms: $a = 2^{2014}$, $b = 2^{402} \cdot 3^4$, $c = 2^{154} \cdot 3^{12}$, $d = 2^{64} \cdot 3^{30}$ or $e = 2^{30} \cdot 3^{12} \cdot 5^4$.

We claim that the smallest number is $2^{30} \cdot 3^{12} \cdot 5^4$. Indeed, $a > e$ since $2^{1984} > 3^{12} \cdot 5^4$; $b > e$ since $2^{372} > 3^8 \cdot 5^4$; $c > e$ for $2^{124} > 5^4$ and $d > e$ for $2^{34} \cdot 3^{18} > 5^4$.