Find positive integers $a, b$ such that $491! = 11^a \cdot 7^b \cdot c $ 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. 
 A: 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


*

*one more factor for $11\cdot 44$

*one more factor for $11\cdot 33$

*one more factor for $11\cdot 22$

*one more factor for $11\cdot 11$
therefore the exponent for $11$ is $48$.
A: Multiples of $11$ are $\{11,22,33,...,484\}$ which are $44$ of them and among them we have multiples of $121$ which are $\{121,242,363,484\}$ which are four of them.
Thus the power of $11$ should be $44+4=48$
With $7$ we have multiples of $7$ as $\{7,14,21,...,490\}$ which are $70$ of them. 
Among them we have multiples of $49$ which are $\{ 49,98,...,490\}$ which are $10$ of them.
Among these we have one multiple of $343$ which is $\{343\}$
Thus the power of $7$ is $$70+10+1$$ which is $81$ 
A: 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:
$$\sum_{k=1}^\infty \left[\dfrac{n}{p^k}\right]$$
