Prime Power in Triple-Factorial

By definition of Multifactorial, $2017!!!= 2017 \times 2014 \times 2011 \times \cdots{ } \cdots{ } \cdots{ }\times 4 \times 1$ .

Let $n$ be any factor in RHS of the last equation. Of course, $n \equiv 1$ $(\mod 3)$ .

Now $n$ contributes a $1$ to the total power of $11$ when $n \equiv 0$ $(\mod 11)$ . But the system of linear congruences with the congruences $n \equiv 1$ $(\mod 3)$ and $n \equiv 0$ $(\mod 11)$ has, by Chinese Remainder Theorem , $n \equiv 22$ $(\mod 33)$ as its solution. As $\lfloor \dfrac{2017}{33}\rfloor=61$ and $2017 \mod 33=4 <22$ , number of such $n$ is $61$ .

Again, $n$ contributes an additional $1$ to the total power of $11$ when $n \equiv 0$ $(\mod 11^2)$ . Generally speaking, $n$ contributes an additional $1$ to the total power of $11$ when $n \equiv 0$ $(\mod 11^k)$ , for $k\geq 1$ .

As $11^4 > 2017$ , we test for $k=1,2$ and $3$ .

For $k=1$ , we've already found the number of such $n$ is $61$ .

Similarly, For $k=2$ , the number of such $n$ is $6$ .

And For $k=3$ , the number of such $n$ is $0$ .

So, $e=61+6+0=\boxed{67}$ .