For every integer $n\ge2$, let $\text{pow}(n)$ be the largest power of the largest prime that divides $n$. For example $\text{pow}(144)=\text{pow}(2^4\cdot3^2)=3^2$. What is the largest integer $m$ such that $2010^m$ divides


$\prod_{n=2}^{5300}\text{pow}(n)$?



$\textbf{(A)}\ 74 \qquad \textbf{(B)}\ 75 \qquad \textbf{(C)}\ 76 \qquad \textbf{(D)}\ 77 \qquad \textbf{(E)}\ 78$
Because 67 is the largest prime factor of 2010, it means that in the prime factorization of $\prod_{n=2}^{5300}\text{pow}(n)$, there'll be $p_1 ^{e_1} \cdot p_2 ^{e_2} \cdot .... 67^x ...$ where $x$ is the desired value we are looking for. Thus, to find this answer, we need to look for the number of times $67$ is incorporated into the giant product.
All numbers $n=67 \cdot x$, given $x = p_1 ^ {e_1} \cdot p_2 ^{e_2} \cdot ... \cdot p_m ^ {e_m}$ such that for any integer $x$ between $1$ and $m$, prime $p_x$ must be less than $67$, contributes a 67 to the product. Considering $67 \cdot 79 < 5300 < 67 \cdot 80$, the possible values of x are $1,2,...,70,72,74,...78$, since $x=71,73,79$ are primes that are greater than 67. However, $\text{pow}\left(67^2\right)$ contributes two $67$s to the product, so we must count it twice. Therefore, the answer is $70 + 1 + 6 = \boxed{77} \Rightarrow \boxed{D}$.