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.

  • 1
    $\begingroup$ $$196-2\cdot97=?$$ $\endgroup$ – lab bhattacharjee May 7 at 15:02

You have found that $400!=3^{196}m$ and $200!=3^{97}p$ where $m,p$ are integers not divisible by $3$. Then $$\frac {400!}{200!^2}=\frac {3^{196}m}{(3^{97}p)^2}=\frac {3^{196}m}{3^{2\cdot97}p^2}=3^2\frac m{p^2}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.