2009
Problem - 3727
If $17!=355687ab8096000$ where $a$ and $b$ are two missing single digits. Find $a$ and $b$.
Because $9\mid 17!$ and $11\mid 17!$, the following must hold:
$$3+5+5+6+8+7+a+b+8+0+9+6+0+0+0\equiv a+b+3\equiv 0\pmod{9}$$
and
$$3-5+5-6+8-7+a-b+8=0+9=6+0-0+0\equiv a-b-2\equiv 0\pmod{11}$$
The only pair of $(a,b)$ that can satisfy the above is $a=4$ and $b=2$.