Question:
Given that $m$ and $n$ are positive integers such that $m\equiv 6\pmod 9$ and $n\equiv 0\pmod 9$, what is the largest integer that $mn$ is necessarily divisible by?

Answer:
If $m\equiv 6\pmod 9$, then we can write $m$ as $9a+6$ for some integer $a$. This is equal to $3(3a+2)$, so $m$ is certainly divisible by $3$.  If $n\equiv 0\pmod 9$, then $n$ is divisible by $9$. Therefore, $mn$ must be divisible by $3\cdot 9 = 27$.

Note that $m$ can be 6 and $n$ can be 9, which gives us $mn = 54$.  Also, $m$ can be 15 and $n$ can be 9, which gives us $mn = 135$.  The gcd of 54 and 135 is 27.

Therefore, the largest integer that $mn$ must be divisible by is $\boxed{27}$.