HMMT February, 2015

Solution:

Answer: $-501$

We claim that
$$p = \frac{1}{5} \frac{6}{10} \frac{11}{15} \frac{16}{20} \cdots \frac{2006}{2010} \frac{2011}{2015}.$$
Let $p_n$ be the probability that, starting with $n$ rocks, the number of rocks left after each round is a multiple of $5$. Indeed, using recursions we have
$$p_{5k} = \frac{p_{5k-5} + p_{5k-10} + \cdots + p_5 + p_0}{5k}$$
for $k \geq 1$. For $k \geq 2$ we replace $k$ with $k-1$, giving us
$$\begin{gathered} p_{5k-5} = \frac{p_{5k-10} + p_{5k-15} + \cdots + p_5 + p_0}{5k-5} \\ \Longrightarrow \\ (5k-5) p_{5k-5} = p_{5k-10} + p_{5k-15} + \cdots + p_5 + p_0 \end{gathered}$$
Substituting this back into the first equation, we have
$$5k p_{5k} = p_{5k-5} + \left(p_{5k-10} + p_{5k-15} + \cdots + p_5 + p_0\right) = p_{5k-5} + (5k-5) p_{5k-5}$$
which gives $p_{5k} = \frac{5k-4}{5k} p_{5k-5}$. Using this equation repeatedly along with the fact that $p_0 = 1$ proves the claim.

Now, the power of $5$ in the denominator is $v_5(2015!) = 403 + 80 + 16 + 3 = 502$, and $5$ does not divide any term in the numerator. Hence $a = -502$. (The sum counts multiples of $5$ plus multiples of $5^2$ plus multiples of $5^3$ and so on; a multiple of $5^n$ but not $5^{n+1}$ is counted exactly $n$ times, as desired.)

Noting that $2015 = 31 \cdot 65$, we found that the numbers divisible by $31$ in the numerator are those of the form $31 + 155k$ where $0 \leq k \leq 12$, including $31^2 = 961$; in the denominator they are of the form $155k$ where $1 \leq k \leq 13$. Hence $b = (13 + 1) - 13 = 1$ where the extra $1$ comes from $31^2$ in the numerator. Thus $a + b = -501$.