Square Cube $5$ th Power-II

First, we must work out what $N$ is.

For $N$ to be as small as possible, its only prime factors have to be $2$ , $3$ , and $5$ .

Let $N = 2^a3^b5^c$ , where $a$ , $b$ , and $c$ are non-negative integers.

We want to make $a$ , $b$ , and $c$ individually as small as possible.

For all conditions to be met:

$lcm(3, 5)|a$ , and $a ≡ -1 ≡ 1 (mod 2)$ .

$lcm(2, 5)|b$ , and $b ≡ -1 ≡ 2 (mod 3)$ .

$lcm(2, 3)|c$ , and $c ≡ -1 ≡ 4 (mod 5)$ .

$lcm(3,5) = 15$ , $lcm(2,5) = 10$ , and $lcm(2,3) = 6$ .

The smallest non-negative integers fulfilling these requirements are

$a = 15$ .

$b = 20$ .

$c = 24$ .

Therefore $N = 2^{15} 3^{20} 5^{24}$ , and $N$ has $(15+1)(20+1)(24+1) = 16*21*25 = 8400$ positive divisors (including $1$ and itself).

Since $\sqrt{2N}$ , $\sqrt[3]{3N}$ , and $\sqrt[5]{5N}$ are positive integers. $N$ must be of the form $2^a \times 3^b \times 5^c$ and we need to find the smallest $a$ , $b$ , and $c$ . Then

$\begin{cases} a+1 \equiv 0 \text{ (mod 2)} \\ a \equiv 0 \text{ (mod 3)} \\ a \equiv 0 \text{ (mod 5)} \end{cases} \implies a \equiv 15m \equiv m \equiv -1 \text{( mod 2)} \implies m = 1 \implies a = 15$

$\begin{cases} b \equiv 0 \text{ (mod 2)} \\ b+1 \equiv 0 \text{ (mod 3)} \\ b \equiv 0 \text{ (mod 5)} \end{cases} \implies b \equiv 10n \equiv n \equiv -1 \text{( mod 3)} \implies n = 2 \implies b = 20$

$\begin{cases} c \equiv 0 \text{ (mod 2)} \\ c \equiv 0 \text{ (mod 3)} \\ c+1 \equiv 0 \text{ (mod 5)} \end{cases} \implies c \equiv 6p \equiv p \equiv -1 \text{( mod 3)} \implies p = 4 \implies c = 24$

Therefore, $N = 2^{15} \times 3^{20} \times 5^{24}$ and the number of divisors is $(15+1)(20+1)(24+1) = \boxed{8400}$ .