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.
Best Answer
Following the hints given in the comments, since $11\cdot 44=484$, for $491!$ the factor $11$ appears at least $44$ times and we need to add
therefore the exponent for $11$ is $48$.