TITLE: Find positive integers $a, b$ such that $491! = 11^a \cdot 7^b \cdot c $
QUESTION [6 upvotes]: The question is to find positive integers $a, b$ such that $491! = 11^a \cdot 7^b \cdot c $ where $c$ is a natural that is not divisible by either $11$ or $7$.
Giving it some thought I believe to have reached something tangible, but not too sure on whether or not it suffices: 
Of course $491! = (491)(490)...(2)(1)$ so any multiple of 11 which is $<491$ will be here. That is, every $11 \cdot k $ for $k\le 45$ $\space$ (for $11*46>491$). So the exponent of $11$ in the prime decomposition of $491!$ will be at least $45$. But we have to account for the numbers which are divisible by powers of $11$. How to do that? Well, we've already established $k \le45$ so there is only one number with such property: $11^2$. Now we just have to sum $44$ (for the numbers divisible only "one time" by $11$) and $2$. That is $46$. 
Now, for the exponent of $7$ the work to be done is analogous. 
It might seem a bit trivial to some of you, but it's bugging not to know whether my reasoning is correct. Thank you in advance for your time and help.

REPLY [2 votes]: For $a$, we first count the number of multiples of $11$, which are $\left[\dfrac{491}{11}\right]=44$. However, the multiples of $11^2$ give a more "$11$", so we need to count again: $\left[\dfrac{491}{11^2}\right]=4$. Therefore, $a=44+4=48$.
Similarly, $b=\left[\dfrac{491}{7}\right]+\left[\dfrac{491}{7^2}\right]+\left[\dfrac{491}{7^3}\right]=70+10+1=81$.
Actually, if you want to find the power of a prime $p$ of the prime factorization of $n!$, you can use this summation:
$$\sum_{k=1}^\infty \left[\dfrac{n}{p^k}\right]$$