Factorial Divisors

Q. For what maximum value of $n$ will the expression $\frac{10200!}{504^n}$ be an integer? I have the solution to this question and I would like you to please go through the solution below. My doubt follows the solution :)

The solution can be found by writing $504 = 2^3 \cdot 3^2 \cdot 7$ and then finding the number of $2^3$s, $3^2$s and $7$s in the numerator, which can be obtained by

Number of $2$s = $\left\lfloor\frac{10200}{2}\right\rfloor + \left\lfloor\frac{10200}{2^2}\right\rfloor + \left\lfloor\frac{10200}{2^3}\right\rfloor + \dots + \left\lfloor\frac{10200}{2^{13}}\right\rfloor= 10192$
where $\left\lfloor\dots\right\rfloor$ is the floor function.

Therefore, the number of $2^3\textrm{s} = \left\lfloor\frac{10192}{3}\right\rfloor = 3397$

$\begin{align}\textrm{Similarly, the number of }3^2\textrm{s} &= 2457\\ \textrm{and the number of }7\textrm{s} & = 1698\end{align}$

The number of factors of $2^3 \cdot 3^2 \cdot 7$ is clearly constrained by the number of $7$s, therefore $n = 1698$.

My question is, whether there is any way I can simply look at the prime factors of the divisor and know which prime factor is going to be the constraining factor? (as $7$ was, in this particular example)