If $p$ is a prime number greater than 2 and $k\in \mathbb{N}$ so that $k < p$, how can I prove that $p\choose k$ is congruent to $0 \mod p$. I'm sorry I know my formatting is rough but I don't know how to format it correctly.
Tell me more
×
Mathematics Stack Exchange is a question and answer site for
people studying math at any level and professionals in related fields. It's 100% free, no registration required.
|
|
|||||
|
|
Hint: Do you know that ${p \choose k}=\frac {p!}{k!(p-k)!}?$ |
|||||||||||
|
|
So, you have: $$\frac{p!}{k!(p-k)!}$$ and you want to prove this is a multiple of $p$. Hint: $n! = n(n-1)!$ |
|||
|
|
|
For $\,1\leq k\leq p-1\,$: $$\binom{p}{k}=\frac{p!}{k!(p-k)!}=\frac{(k+1)(k+2)\cdot\ldots\cdot (p-1)}{1\cdot 2\cdot\ldots\cdot (p-k)}\cdot\;p$$ Since the binomial coefficient is an integer and all the factors both in the numerator and in the denominator above are smaller than $\,p\,$, we're done |
|||
|
|
