Yes, you are right. (Happy)

Printable View

- Jul 11th 2010, 02:56 PMBen92
Yes, you are right. (Happy)

- Jul 11th 2010, 06:10 PMmelese
Hello chiph588@, I think I've found what you're looking for, namely a sufficient and necessary condition of the divisibility of $\displaystyle \binom{2n}{n} $ by a prime $\displaystyle p $ in general!

A prime $\displaystyle p $ divides $\displaystyle \binom{2n}{n} $ if and only if the fractional part $\displaystyle \left \{ \frac{n}{p^k} \right \}\geq\frac{1}{2} $ for at least one $\displaystyle k\in \mathbb{N}$. In particular, when $\displaystyle k=1 $.

For example, $\displaystyle 3\not|\binom{8}{4}$ and indeed $\displaystyle \left \{ \frac{4}{3^1} \right \}<\frac{1}{2}$; $\displaystyle \left \{ \frac{4}{3^2} \right \}<\frac{1}{2}$; and so on...

Another example, the one that you gave here is $\displaystyle 3\mid\binom{14}{7}

$ and as you noted $\displaystyle \left\{\frac73\right\}=\frac13$, but if you continue $\displaystyle \left\{\frac{7}{3^2}\right\}\geq\frac12$. - Jul 11th 2010, 06:12 PMchiph588@
- Jul 11th 2010, 08:43 PMBen92
Me too.