Calculate the number of zeros at the end of 2015!. 

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As $10=2 \times 5$, the number of zeros at the end of 2015! is $\min \left(v_{2}(2015!), v_{5}(2015!)\right)$. However, there are more factors of 2 than factors of 5. Therefore, we calculate $v_{5}(2015!)$ using Legendre's formula:

$$
v_{5}(2015!)=\sum_{k=1}^{+\infty}\left[\frac{2015}{5^{k}}\right]=502
$$

## - Chinese Remainder Theorem -

Theorem 146 (Chinese Remainder Theorem). Let $a_{1}, \ldots, a_{d}$ be integers and $N_{1}, \ldots, N_{d}$ be strictly positive integers that are pairwise coprime. Then there exists an integer $a$ such that the system of congruences:

$$
\left\{\begin{aligned}
x & \equiv a_{1}\left[N_{1}\right] \\
& \vdots \\
x & \equiv a_{d}\left[N_{d}\right]
\end{aligned}\right.
$$

is equivalent to the single congruence $x \equiv a\left[N_{1} \ldots N_{d}\right]$

Proof. First, note that it suffices to prove the theorem when $k=2$. A direct induction will then allow us to have it in its full generality. We seek to solve the equation

$$
\begin{cases}x & \equiv a_{1}\left[N_{1}\right] \\ x & \equiv a_{2}\left[N_{2}\right]\end{cases}
$$

The first condition ensures the existence of an integer $q$ such that $x=a_{1}+q N_{1}$, and the second congruence can then be written as:

$$
a_{1}+q N_{1} \equiv a_{2}\left[N_{2}\right]
$$

which provides:

$$
q \equiv\left(a_{2}-a_{1}\right) N_{1}^{\prime}\left[N_{2}\right]
$$

where $N_{1}^{\prime}$ denotes an inverse of $N_{1}$ modulo $N_{2}$, which exists because $N_{1}$ and $N_{2}$ are assumed to be coprime. Thus, by setting $a=a_{1}+\left(a_{2}-a_{1}\right) N_{1}^{\prime} N_{1}$, we obtain $x \equiv a\left[N_{1} N_{2}\right]$.

The converse is immediate.