Legendre would be proud of you

First, let's see how many trailing zeroes there are in $31!$ . For this, we look for pairs of numbers in $31!$ whoose product is divisible by 10.

The last six trailing zeroes are given in the problem, so one is missing, which means $\boxed{Y = 0}$ .

In the remaining factors, there are still enough even numbers to make the product divisible by 16. The divisibility rule for 16 is that the numbrt formed by the last 4 digits is divisible by 16. This number is $628X = 392 \cdot 16 + 8 + X$ . To make it divisible by 16, X has to cancel out the 8, so $\boxed{X = 8}$ .

Combining both results, the answer is $\overline{XY} = \boxed{80}$ .

We can factorize a factorial as $n! = \displaystyle \prod_{k=1}^\infty p_k^{q_k}$ , where $p_k$ is the $k$ th prime and $q_k$ is the highest power of $p_k$ . If $p_k$ is not a factor of $n!$ , then $q_k = 0$ . We can find $q_k$ by $q_k = \displaystyle \sum_{j=1}^\infty \left \lfloor \frac n{p_k^j} \right \rfloor$ , where $\lfloor \cdot \rfloor$ denotes the floor function. For $n=31$ , we get:

$\begin{aligned} 31! & = {\color{#3D99F6}2^{26}}\cdot3^{14}\cdot {\color{#3D99F6}5^{7}} \cdot7^{4}\cdot11^{2}\cdot13^{2}\cdot17^{1}\cdot19^{1}\cdot23^{1}\cdot29^{1}\cdot31^{1} & \small \color{#3D99F6} \text{As }2^7 \cdot 5^7 = 10^7 \\ & = {\color{#3D99F6}2^{19}}\cdot3^{14} \cdot7^{4}\cdot11^{2}\cdot13^{2}\cdot17^{1}\cdot19^{1}\cdot23^{1}\cdot29^{1}\cdot31^{1} \cdot \color{#3D99F6} 10^7 & \small \color{#3D99F6} \text{or 7 trialing 0's} \end{aligned}$

This means that $Y=0$ . Now let $N = \dfrac {31!}{10^7}$ . Then $X = N \bmod 10$ .

$\begin{aligned} N & \equiv 2^{19}\cdot3^{14} \cdot7^{4}\cdot11^{2}\cdot13^{2}\cdot 17 \cdot19 \cdot 23 \cdot 29\cdot 31 \text{ (mod 10)} \\ & \equiv (16^4 \cdot 8) \cdot 9^7 \cdot 49^2 \cdot 1^2 \cdot 3^2 \cdot 7 \cdot 9 \cdot 3 \cdot 9 \cdot 1 \text{ (mod 10)} \\ & \equiv (6\cdot 8) (10-1)^7 (50-1)^2 (1)(-1)(7) (-1) (3) (-1)(1) \text{ (mod 10)} \\ & \equiv (8) (-1)(-1)(-1)(-1) \text{ (mod 10)} \\ & \equiv 8 \text{ (mod 10)} \end{aligned}$

Implying $X=8$ and $\overline{XY} = \boxed{80}$ .