Square Cube 5th Power

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 $a ≡ 1 (mod 2)$ .

$lcm(2, 5)|b$ , and $b ≡ 1 (mod 3)$ .

$lcm(2, 3)|c$ , and $c ≡ 1 (mod 5)$ .

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

The smallest non-negative integers fulfilling these requirements are

$a = 15$ .

$b = 10$ .

$c = 6$ .

Therefore $N = 2^{15} 3^{10} 5^6$ , and $N$ has $(15+1)(10+1)(6+1) = 16*11*7 = 1232$ positive divisors (including $1$ and itself).