Progetto Olimpiadi della Matematica

Solution:

The answer is (A). We know that $y$ is the largest exponent such that $10^{y}$ divides $2013!$. Thus $y$ is the minimum between the number of factors $2$ and factors $5$ that occur in the factorization of $2013!$. Since the number of factors two is clearly greater than the number of factors $5$, $y$ is also the number of factors $5$ in the factorization of $2013!$.

Similarly, the number of zeros with which $2000!$ written in base $5$ ends is simply the number of factors $5$ that appear in the factorization of $2000!$.

Let us then write $2000! = 5^{x} \cdot a$, $2013! = 5^{y} \cdot b$ for two integers $a, b$ in whose factorization no factors $5$ appear.

Let us now observe that $\frac{2013!}{2000!} = 5^{y-x} \cdot \frac{a}{b}$ is an integer, because it coincides with $2001 \cdot 2002 \cdots 2013$.

It follows that $\frac{a}{b}$ is an integer with no factors five, and that $y-x$ is the number of factors $5$ in the factorization of $2001 \cdot 2002 \cdots 2013$. This latter quantity is easy to compute: the only multiples of $5$ in this product are $2005$, $2010$, and both contribute exactly one factor five (not being divisible by $25$), so $y-x=2$.