More fun in 2015, Part 41

We can group the divisors of a square-free integer $n>1$ into (unordered) pairs $d$ and $n/d$ , with $\phi(d)\phi(n/d)=\phi(n)$ since $\phi$ is multiplicative. If $\tau(n)$ is the number of divisors of $n$ , then there are $\tau(n)/2$ such pairs, so that $\prod_{d|n}\phi(d)=\phi(n)^{\tau(n)/2}$ and $L=\tau(n)/2$ . Note that $\tau(n)=2^m$ if $n$ has $m$ prime factors.

In the case of $n=2014*2015$ we have $m=6$ prime factors so that $L=2^5=\boxed{32}$ .

This solution will focus on n squarefree. Since n is squarefree, we may let $\displaystyle n = \prod_{i=1}^{\omega(n)} p_i$ , where $\omega(n)$ is the number of distinct prime factors of n.

"Zoning in" on a single factor $p_1$ , we can ask how many divisors of n have $p_1$ as a factor. If a divisor has m factors, one of which is $p_1$ , then we can choose m-1 other factors in $\dbinom{\omega(n)-1}{m-1}$ ways. It is then easy to conclude that, in total, there are $\dbinom{\omega(n)-1}{1-1} + \dbinom{\omega(n)-1}{2-1} + \ldots + \dbinom{\omega(n)-1}{\omega(n)-1} = 2^{\omega(n)-1}$ divisors of n with $p_1$ ; an identical argument follows for all other $p_i$ .

Using the fact that $\phi$ is multiplicative, we may simply separate the factors of each divisor. Following this, we may observe that

$\begin{aligned} \prod_{d|n} \phi(d) &= \prod_{i=1}^{\omega(n)} \phi(p_i)^{2^{\omega(n)-1}} \\ &= \left[ \prod_{i=1}^{\omega(n)} \phi(p_i) \right]^{2^{\omega(n)-1}} \\ &= \phi(n)^{2^{\omega(n)-1}}. \end{aligned}$

This implies, then, that (for $\phi(n) \neq 1$ ),

$\begin{aligned} \log_{\phi(n)} \left( \prod_{d|n} \phi(d) \right) &= \log_{\phi(n)} \phi(n)^{2^{\omega(n)-1}} \\ &= 2^{\omega(n)-1}. \end{aligned}$

Here, $\omega(2014 \times 2015) = 6$ so $2^{\omega(n)-1} = \boxed{32}$ .