9. Let $N$ be the smallest positive integer such that $N / 15$ is a perfect square, $N / 10$ is a perfect cube, and $N / 6$ is a perfect fifth power. Find the number of positive divisors of $N / 30$.
Answer: 8400
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 $16 \cdot 21 \cdot 25=8400$.