For how many integers $a$ is $\frac{2^{10} \cdot 3 ^8 \cdot 5^6}{a^4}$ an integer?

Question

In Mathleague $11316$ Target #$4$, the question is:

For how many integers $a$ is $$\frac{2^{10} \cdot 3 ^8 \cdot 5^6}{a^4}$$ an integer?

Answer 1

Clearly the number $$ q =\frac{2^{10} 3^8 5^6 }{a^4} = \left( \frac{2^{5} 3^4 5^3 }{a^2} \right)^2 $$ is a perfect square, and so $$ \sqrt{q} = \frac{2^{5} 3^4 5^3 }{a^2} $$ is rational. Now in order for this to be an integer, the denominator must divide the numerator. Therefore $a^2$ must divide $2^53^45^3$. Now the greatest perfect square dividing $2^53^45^3$ is $2^4 3^4 5^2$, and so $a^2$ can be any divisor of this number. Therefore $a$ can be any integer divisor of its square root, $2^2 3^2 5$. There are $3(3)(2)=18$ positive divisors of this number, and so there are $36$ signed possibilities for $a$. So the answer is 36.

Answer 2

Hint: Factor $a$ into primes; it's evident that these primes must be $2$, $3$, and $5$ (why?). Then $a = 2^d \cdot 3^b \cdot 5^c$, so that

$$a^4 = 2^{4d} \cdot 3^{4b} \cdot 5^{4c}$$

What conditions on $d, b, c$ are necessary and sufficient to make this a divisor of $2^{10} \cdot 3^8 \cdot 5^6$?