# Tagged Questions

56 views

### The binomial coefficients $\binom{n}{ p}$ are divisible by a prime $p$ only if $n$ is a power of $p.$

I'm looking for a "high school / undergraduate" demonstration for the: All the binomial coefficients $\binom{n}{i}=\frac{n!}{i!\cdot (n-i)!}$ for all i, $0\lt i \lt n$, are divisible by a prime $p$ ...
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### Divisibility of binomial coefficient by prime power - Kummer's theorem

Let's say we have binomial coefficient $\binom{n}{m}$. And we need to find the greatest power of prime $p$ that divides it. Usually Kummer's theorem is stated in terms of the number of carries you ...
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### Proofs from the Book - need quick explanation

I've been recently reading this amazing book, namely the chapter on Bertrand's postulate - that for every $n\geq1$ there is a prime $p$ such that $n<p\leq2n$. As an intermediate result, they prove ...
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### On Elements of $p$th Row in n Pascal's Triangle (For Prime $p$)

If $p$ is a prime number, in Pascal's triangle all the terms in the $p$th row - except the 1s - are multiples of $p$ . It's easy to prove this property using the formula for $\binom{p}{k}$. Is there ...
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### Binomial coefficient modulo composite number [closed]

For a prime number n , ${n \choose r} \mod n$ is 0 for all 0 < r < n If n is composite I want to know for which values of r the value of ${n \choose r} \mod n$ becomes 0 (for r in ...
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### Proving that $n\mid(nCr)$ for all $r$ ($1 \leq r \leq n-1$), only if $n$ is prime

I'm trying to prove that $n\mid(nCr)$ for all $r$ ($1 \leq r \leq n-1$) if and only if $n$ is prime. Now proving that if $n$ is prime then $n\mid(nCr)$ is pretty easy, but how would you go about ...
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### Bertrand's postulate proof

Regarding http://michaelnielsen.org/polymath1/index.php?title=Bertrand%27s_postulate I think the last inequality should be $4^{n/3}\le(2n+1)(2n)^{\sqrt{2n}}$. But even when the RHS is decreased from ...
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### Binomial theorem for prime exponent

Could you explain to me why for prime $p$ we have the following? $$(x+y)^p - (x^p + y^p)= x^p + \binom{p}{1}x^{p-1}y + \binom{p}{2}x^{p-2}y^2 + \binom{p}{p-1}xy^{p-1} + y^p.$$ I found it here: ...