Find the smallest value of $n$ so that the greater potency of $5$ which divides $n!$ is $5^{84}$. What are the other numbers that enjoy this property?
I thought I would put together an equation of the type $$E_5(n!)=84\\84=\left[\frac{n}{5} \right]+\left[\frac{n}{5^2} \right]+\left[\frac{n}{5^3} \right]+\;...$$ Only I do not see how to proceed, or if this path will take me where I want, someone could help me?
You have
$$E_p(n!) = \sum_{k=1}^\infty \left\lfloor \frac{n}{p^k}\right\rfloor < \sum_{k=1}^\infty \frac{n}{p^k} = \frac{n}{p} \sum_{r=0}^\infty \frac{1}{p^r} = \frac{n}{p}\frac{1}{1-\frac1p} = \frac{n}{p-1},$$
and the difference is small(ish). The first sum is actually finite, all terms for $k > \frac{\log n}{\log p}$ are $0$, writing it as an infinite sum is just convenient.
So to have $E_p(n!) = m$, a ball-park estimate is $n \approx (p-1)\cdot m$, and we must have $n > (p-1)\cdot m$. For $p=5$ and $m = 84$, that yields $n \approx 4\cdot 84 = 336$.
The smallest $n$ will of course be a multiple of $5$, so let's check $n = 340$:
$$E_5(340!) = 68 + 13 + 2 = 83,$$
which is a little too small, we need one more power, so $n = 345$ is it.