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I was working through a problem set for number theory and needed some help with this problem:

Find the largest integer $k$ for which $3^k$ divides $400\choose 200$.

I know this will reduce to $\frac{400!}{(200!)^2}$ and I found that the largest power of $3$ that divides $400!$ is $196$ and the largest power that divides $200!$ is $97$ but I don't know how to put them together.

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    $\begingroup$ $$196-2\cdot97=?$$ $\endgroup$ – lab bhattacharjee May 7 at 15:02
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You have found that $400!=3^{196}m$ and $200!=3^{97}p$ where $m,p$ are integers not divisible by $3$. Then $$\frac {400!}{200!^2}=\frac {3^{196}m}{(3^{97}p)^2}=\frac {3^{196}m}{3^{2\cdot97}p^2}=3^2\frac m{p^2}$$

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