Problem: Let $\varphi(n)$ denote the number of positive integers
less than $n$ that are relatively prime to $n$.
Prove that there exists a positive integer $m$ for which the equation
$\varphi(n) = m$ has at least $2015$ solutions in $n$.

Solution: Here are two explicit solutions.

\paragraph{First solution with ad-hoc subsets, by Evan Chen.}
I consider the following eleven prime numbers:
\[ S = \left\{ 11, 13, 17, 19, 29, 31, 37, 41, 43, 61, 71 \right\}. \]
This has the property that for any $p \in S$,
all prime factors of $p-1$ are one digit.

Let $N = (210)^{\text{billion}}$,
and consider $M = \varphi\left( N \right)$.
For any subset $T \subset S$, we have
\[ M = \varphi\left(
  \frac{N}{\prod_{p\in T} (p-1)} \prod_{p \in T} p \right). \]
Since $2^{\left\lvert S \right\rvert} > 2015$ we're done.
\begin{remark*}
This solution is motivated by the deep fact that
$\varphi(11 \cdot 1000) = \varphi(10 \cdot 1000)$,
for example.
\end{remark*}

\paragraph{Second solution with smallest primes, by Yang Liu.}
Let $2 = p_1 < p_2 < \dots < p_{2015}$
be the \emph{smallest} $2015$ primes.
Then the $2015$ numbers
\begin{align*}
  n_1 &= (p_1-1) p_2 \dots p_{2015} \\
  n_2 &= p_1 (p_2-1) \dots p_{2015} \\
  &\vdotswithin= \\
  n_{2015} &= p_1 p_2 \dots \left( p_{2015}-1 \right)
\end{align*}
all have the same phi value, namely
\[ \varphi(p_1 p_2 \dots p_{2015}) = \prod_{i=1}^{2015} (p_i-1). \]