smc

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 $70+1+6=\boxed{77}\Rightarrow \boxed{77}$.