Simple but sweet

From $n=2a^2=3b^3=5c^5$ , we note that since $n, a, b, c \in \mathbb N$ , then $n$ must have the prime factors of 2, 3 and 5. To be the smallest, $n$ must not have other prime factors. Therefore, we can assume $n = 2^x3^y5^z$ , where $x, y, z \in \mathbb N$ . Then we have $2a^2 = 2^x3^y5^z$ $\implies a^2 = 2^{x-1}3^y5^z$ $\implies a = 2^{\frac {x-1}2}3^\frac y2 5^\frac z2$ . Similarly, we have:

$\large \begin{cases} a = 2^{\frac {x-1}2}3^\frac y2 5^\frac z2 \\ b = 2^\frac x3 3^{\frac {y-1}3}5^\frac z3 \\ c = 2^\frac x5 3^\frac y5 5^{\frac {z-1}5} \end{cases}$

For $a$ , $b$ and $c$ to be positive integers $\dfrac {x-1}2$ , $\dfrac x3$ and $\dfrac x5$ must be positive integers too. Therefore $x-1 \equiv 0 \text{ (mod 2)}$ or $x \equiv 1 \text{ (mod 2)}$ , $x \equiv 0 \text{ (mod 3)}$ and $x \equiv 0 \text{ (mod 5)}$ . Implies that $x \equiv 15 g \equiv 1 \text{ (mod 2)}$ and the smallest $x=15$ . Similarly,

$\begin{cases} x \equiv 1 \text{ (mod 2)} \equiv 0 \text{ (mod 3)} \equiv 0 \text{ (mod 5)} & \implies x \equiv 15g \equiv 1 \text{ (mod 2)} & \implies x = 15 \\ y \equiv 0 \text{ (mod 2)} \equiv 1 \text{ (mod 3)} \equiv 0 \text{ (mod 5)} & \implies y \equiv 10h \equiv 1 \text{ (mod 3)} & \implies y = 10 \\ z \equiv 0 \text{ (mod 2)} \equiv 0 \text{ (mod 3)} \equiv 1 \text{ (mod 5)} & \implies z \equiv 6k \equiv 1 \text{ (mod 5)} & \implies z = 6 \end{cases}$

$\implies n = 2^{15} 3^{10} 5^6$ . Therefore, the answer $=(15+1)(10+1)(6+1) = \boxed{1232}$

It is obvious that n must be divisible by 2, 3 and 5 because a, b, c are positive integers. We have to find the least possible value of n and if $n = 2^d \cdot 3^e \cdot 5^f · q$ satisfies, then n = $2^d$ · $3^e$ · $5^f$ satisfies too. Because of this n have to be a form of n = $2^d$ · $3^e$ · $5^f$ . According to the problem,

$2^{d-1} 3^e5^f = a^2$ ,

$2^d$ $3^e$ $^-$ $^1$ $5^f$ = $b^3$ ,

$2^d$ $3^e$ $5^f$ $^-$ $^1$ = $c^5$ .

Because $a^2$ is a square 2|d-1, 2|e and 2|f. In the same way: 3|d, 3|e-1, 3|f and 5|d, 5|e, 5|f-1

Now we can find the exact values of d, e and f. The least odd number, which is divisible by 3 and 5 is 3·5 = 15, thus d = 15. The least number, which is divisible by 2 and 5 is 2·5 = 10, and 3|10-1, thus e = 10. The least number, which is divisible by 2 and 3 is 2·3 = 6, and 5|6-1, thus f = 6.

And finally, we have got n = $2^1$ $^5$ $3^1$ $^0$ $5^6$ , so the answer is ((15+1)(10+1)(6+1) = 1232 .