jmc

If we want the largest such $n$, we must count the number of factors of 3 in the prime factorization of $200!$. The number of multiples of 3 is $\lfloor \frac{200}{3}\rfloor$. (The notation $\lfloor x\rfloor$ means the greatest integer less than or equal to $x$, so basically, $\lfloor \frac{200}{3}\rfloor$ means ``divide 200 by 3 and round down.") However, these are not the only factors of 3 in the product! Multiples of 9 in the product contribute an extra factor of 3. There are $\lfloor \frac{200}{9}\rfloor$ of these, so we add this number to the $\lfloor \frac{200}{3}\rfloor$ factors of 3 we found from counting multiples of 3. Similarly, we must add in multiples of 27 once and multiples of 81.

The sum is $\lfloor \frac{200}{3}\rfloor +\lfloor \frac{200}{9}\rfloor +\lfloor \frac{200}{27}\rfloor +\lfloor \frac{200}{81}\rfloor =66+22+7+2=\boxed{97}$.