Factorial Fragments

Maxwell from Samuel Gilbert Public School in Australia has got us started on this problem by noticing two things:

$\log{n!}=\log{n}+\log{(n-1)}+\log{(n-2)}+....+\log{2}+\log{1}$
and
$\log_a{n!}=m+\log_a{\frac{n!}{a^m}}$

The latter observation allows us to see that the integer term will always be the highest power of $a$ that goes into $n!$, so in particular there may not always be an integer term if $a$ is not a factor of $n!$.

Thomas from BHASVIC Sixth Form College in Brighton has furthered this arguement:
If $a$ does not divide $n!$, then there cannot be any integer term.

He also managed to solve the equation
$$\log n!= a \log 2 + \square \log 3 + 3 \log 5 + \square \log 7 + \log p + \log q$$
for $p < q$ both primes greater than $7$ and $a$ not prime.

First use fact that $n!$ has every integer less than and equal to $n$ as a factor. This means that $p$ and $q$ must be the next two primes after $7$, since otherwise these would be missing from the sum, even though they are factors of $n!$ (obviously need $n \geq q$). So $p=11$, $q=13$, and we must have $n \geq 13$.

We must also have $n<17$ since this prime does not appear in the sum, and $n \geq 15$ since we need $5^3$ to divide $n$, which leaves only two possibilities, $n=15$ and $n=16$.

A quick look at the $\log 2$ term tells us that for $n=15$ we get $a=11$ which is prime, but for $n=16$ we have $a=15$, so $n=16$ is the only possibility. Now calculating the empty boxes, get that $\log 16! = 15 \log 2 + 6 \log 3 + 3 \log 5 + 2 \log 7 + \log 11+ \log 13$.

If you found thinking about this problem interesting, you might like to look up an approximation called Stirling's formula:
$$\log n! \sim n\log n - n$$