Philippine Mathematical Olympiad

Solution:

$N$ must be of the form $N = 2^{m} 3^{n} 5^{p}$ for some nonnegative integers $m, n, p$.

Since $N / 15 = 2^{m} 3^{n-1} 5^{p-1}$ is a perfect square, we have $m \equiv 0 \bmod 2$ and $n \equiv p \equiv 1 \bmod 2$.

Since $N / 10 = 2^{m-1} 3^{n} 5^{p-1}$ is a perfect cube, we have $n \equiv 0 \bmod 3$ and $m \equiv p \equiv 1 \bmod 3$.

Since $N / 6 = 2^{m-1} 3^{n-1} 5^{p}$ is a perfect fifth power, we have $p \equiv 0 \bmod 5$ and $m \equiv n \equiv 1 \bmod 5$.

By Chinese remainder theorem, we get $m \equiv 16 \bmod 30$, $n \equiv 21 \bmod 30$ and $p \equiv 25 \bmod 30$.

Since $N$ is as small as possible, we take $m = 16$, $n = 21$, $p = 25$, so $N = 2^{16} 3^{21} 5^{25}$.

Thus the number of positive divisors of $N / 30 = 2^{15} 3^{20} 5^{24}$ is $16 \cdot 21 \cdot 25 = 8400$.