Dissecting the monster: (2016)

For $\phi$ from $45$ to $2016$ , the value of

$\large \lfloor \log_{\phi}{2016} \rfloor$

is $1$ . So, it's a pretty good bet that the answer is $1$ .

Let $x_{1} + x_{2} + x_{3} + x_{4} + \cdots + x_{n} = 2016$

Using, $AM \geq GM$

$\frac{x_{1} + x_{2} + x_{3} + x_{4} + \cdots + x_{n}}{n} \geq (x_{1}x_{2}x_{3}x_{4} \cdots x_{n})^{\frac{1}{n}}$

$\large x_{1}x_{2}x_{3}x_{4} \cdots x_{n} \leq (\frac{x_{1} + x_{2} + x_{3} + x_{4} + \cdots + x_{n}}{n})^{n}$

Thus, we see that maximum value of $\large x_{1}x_{2}x_{3}x_{4} \cdots x_{n}$ is obtained at, $x_{1} = x_{2} = x_{3} = x_{4} = \cdots = x_{n}$

But, here we notice that, $\large (\frac{x_{1} + x_{2} + x_{3} + x_{4} + \cdots + x_{n}}{n})^{n}$ is a discrete function of $n$ .

To arrive at some possible neighbourhood, first we need to make it a continuous function.

Thus by changing the variable from $n$ to $x$ .

$f(x) = (\frac{2016}{x})^{x}$

$f'(x) = 0$

$\therefore (\frac{2016}{x})^{x} . {log(\frac{2016}{x}) - 1} = 0$

$x = \frac{2016}{e}$

Thus the nearest integer comes out to be: $\boxed{742}$