Max value of $n$ for which $3^n\mid(80!)!$ 
Calculate the max value of $n$ for which $(80!)!$ is divisible by $3^n$.

My Attempt:
The exponent of prime factor $p$ in $(n!)$ is given as
$$
v_p(n!) = \left\lfloor \frac{n}{p}\right\rfloor + \left\lfloor \frac{n}{p^2}\right\rfloor+\left\lfloor \frac{n}{p^3}\right\rfloor+\left\lfloor \frac{n}{p^4}\right\rfloor+\cdots
$$
So the exponent of prime factor $3$ in $(80!)$ is given as
$$
v_3(80!) = \left\lfloor \frac{80}{3}\right\rfloor +\left\lfloor \frac{80}{3^2}\right\rfloor+\left\lfloor \frac{80}{3^3}\right\rfloor+\left\lfloor \frac{80}{3^4}\right\rfloor+\cdots=36
$$
But I do not understand how to calculate exponent of $3$ in $(80!)!$.
 A: The exact answer will be $ v_3 ( (80!)!) = \lfloor \frac{80!}{3} \rfloor + \lfloor \frac{80!}{3^2} \rfloor + \ldots $.
If we ignore the floor function, and take the sum of the geometric progression to infinity, the answer would simply be 
$$v_3 ((80!)!) \approx \frac{ 80!}{3} + \frac{80!}{3^2} + \ldots = 80! \times \frac{ \frac{1}{3} } { \frac{2}{3} } = \frac{80!}{2}.$$
Since you already calculated that $ v_3 ( 80!) = 36 $, we know that the first 36 terms of the summation are integers, and the rest of the terms will thus contribute a very small error. In fact, we can hunt down the exact value of this error, by looking at the base 3 representation of $80!$.
Claim: $v_3 (n!) = \frac{n}{2} - R$ where $R$ is half the digit sum of $n$ in base 3.
Proof: This follows immediately by looking at the overestimation in each digit, which is $(0.1111111\ldots)_3 = \frac{1}{2}$. Hence the total overestimation is half the digit sum in base 3. $_\square$
As an explicit example, if we look at $v_3 (80!) = (222)_3 + (22)_3 + (2)_3 $, we have over approximated by $2 \times \frac{1}{2} + 2 \times \frac{1}{2} + 2 \times \frac{1}{2} + 2 \times \frac{1}{2} = 4$. A quick check shows that $ \frac{80}{2} - 4 = 36$, which agrees with your calculation.
It remains to show that $80!$ in base 3 has a digit sum of 220 (I don't know of an immediate way to do this), which would give you Daniel's result that $v_3 (80!) = \frac{80!}{2} - 110$.

Of course, this easily generalizes to other (prime) values.
