How to find the largest prime number where 2012! in the base of the number has at least that many trailing zeros.

Question

Find the largest prime number $p$ such that when $2012!$ is written in base $p$, it has at least $p$ trailing zeroes.

I know that $2012!$ has $501$ trailing zeros. I don't know what to do next or how to use that information.

Answer 1

The last $k$ digits of a number in base $p$ represent its value modulo $p^{k}$. The last $p$ digits in base $p$ are all zero iff the number is divisible by $p^p$. Now, for any prime $p$, the number of powers of $p$ that divide $N!$ is given by $$ \left\lfloor\frac{N}{p}\right\rfloor+\left\lfloor\frac{N}{p^2}\right\rfloor+\left\lfloor\frac{N}{p^3}\right\rfloor+\ldots $$ So you want the largest prime $p$ such that $$ \left\lfloor\frac{N}{p}\right\rfloor+\left\lfloor\frac{N}{p^2}\right\rfloor+\left\lfloor\frac{N}{p^3}\right\rfloor+\ldots \ge p, $$ with $N=2021$. Looks like this will generally be $p \approx \sqrt{N}$, and here it's easy to check that $p=43$ is correct.