Proving that $n\mid(nCr)$ for all $r$ ($1 \leq r \leq n-1$), only if $n$ is prime

Question

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 proving that $n\mid(nCr)$ only if $n$ is prime?

Could you show that if $n$ is not prime, then there exists an $r$ such that $n$ does not divide $nCr$? If so how would you go about doing that? I thought it would be easy but then I realized that for some $r, n\mid(nCr)$ even if $n$ isn't prime which made things a little bit more complicated.

I've managed to show that $n\mid(n-1)!$ for non-primes greater than or equal to $6$, which means that despite $n$ always canceling out with a product of factors in $r!(n-r)!$ in some cases, there will be another pair of factors that cancel out with the $n$ in $(n-1)!$, and sometimes there won't be. Maybe there's some kind of pattern but I can't find it unfortunately.

(I'd prefer a method that doesn't use undergraduate and above level maths)

Jyrki Lahtonen · Answer 1 · 2013-08-21T18:23:23.830

6

Assume first that $n$ is not a power of a prime. Let $p$ be a prime factor of $n$, and let $p^k$ be its highest power dividing $n$, i.e. $p^k\mid n$, $p^{k+1}\nmid n$. Then it follows from Lucas' theorem that $$ p\nmid{n\choose p^k}, $$ in other words $r=p^k$ is the example you asked for.

Assume then that $n=p^k$ for some prime $p$ and integer $k>1$. Then it is not hard to show that $$ p^2\nmid{n\choose p^{k-1}}. $$ This settles the prime power case.

edited Aug 21 '13 at 18:23

answered Aug 21 '13 at 18:17

Jyrki Lahtonen

133,153

Typo in the first sentence, "... is not a power of a prime". – Daniel Fischer Aug 21 '13 at 18:20
What does Lucas' theorem say? – user2566092 Aug 21 '13 at 18:22
For a statement and proof Lucas' theorem see e.g. here. The latter claim is similar in spirit. A result partially generalizing Lucas theorem tells that the power of $p$ dividing $nCr$ is equal to the number of "carry digits" that you get when you calculate $r+(n-r)=n$ in base-$p$ with the grade school pencil & paper method. – Jyrki Lahtonen Aug 21 '13 at 18:23
Thanks, @Daniel. Indeed, there was a typo. – Jyrki Lahtonen Aug 21 '13 at 18:23
For a proof of the latter fact see Arturo Magidin's answer here. He has dug out that the result is due to Kummer. – Jyrki Lahtonen Aug 21 '13 at 18:29

user84413 · Accepted Answer · 2013-08-22T19:28:58.710

3

Suppose n is not prime, and let $n=mp$ where $p$ is a prime. Then $n\not|\binom{n}{p}$, since $$\binom{n}{p}=n \bigg[\frac{(n-1)!}{p!(n-p)!}\bigg]=n\bigg[\frac{(mp-1)!}{p!((m-1)p)!}\bigg]=n\bigg[\frac{(mp-1)(mp-2)(mp-3)\cdots((m-1)p+1)}{p!}\bigg],$$ where the expression in brackets is not an integer since $(mp-1)(mp-2)(mp-3)\cdots((m-1)p+1)$ is not divisible by $p$, while $p!$ is.

edited Aug 22 '13 at 19:28

answered Aug 21 '13 at 18:35

user84413

27,211

Hmm. If $mp-1\ge p^2$ (or $m\ge p$) then the number of times $p$ is a factor in the numerator is higher than $m-1$. The same thing then happens in the denominator, so may be this still works? – Jyrki Lahtonen Aug 21 '13 at 18:45
This does work! The key is that there are no multiples of $p$ in the interval $((m-1)p,mp-1]$. Well done. This is simpler than my suggestion. – Jyrki Lahtonen Aug 21 '13 at 18:52
Thanks for your comments, and you are right that my count of the factors of p in the top and bottom isn't correct. (I was counting multiples, not factors.) I guess the best way to correct my answer would be to observe, as you did, that $(mp-1)!$ and $(mp-p)!$ have the same number of factors of p. – user84413 Aug 21 '13 at 19:24
1

@user84899 I was mistakenly counting multiples rather than factors, but to find the number of factors of p in $n!$, you can take $\lfloor{\frac{n}{p}}\rfloor+\lfloor{\frac{n}{p^2}}\rfloor+\lfloor{\frac{n}{p^3}}\rfloor+\cdots$. – user84413 Aug 21 '13 at 19:40

score 1 · Answer 3 · edited Oct 12 '15 at 05:16

1

We have to prove that n|ncr. now, $nCr=n \cdot(n-1)\cdot(n-2)\cdot(n-3)....\cfrac{n+r-1}{r!}$ Since n is prime and $n \gt r$ then no factors of $r!$ divide $n$. thus $n$ does not cancel.so n must divide ncr when n is prime.

edited Oct 12 '15 at 05:16

BLAZE

8,458

answered Oct 12 '15 at 04:46

Biswajit debnath

19

dionysus · Answer 4 · 2022-02-12T11:54:44.757

I am surprised that people are posting complex proofs of a theorem which is easily provable by high school math. Here is an example.

"if n is prime then n∣(nCr) is pretty easy". right, so we will skip this one. We will just prove the second part, i.e. if n is composite, then $ n \nmid nCr $ for certain r's.

In fact, those r's are all the prime factors of n.

Let r = p where p is a prime factor of n. so n = mp where m is some int.

$ nCp = \frac {n(n-1)(n-2)..(n-p+1)}{p(p-1)(p-2)...1} = (\frac {n}{p}) \frac{(n-1).(n-2)...(n-p+1)}{(p-1)(p-2)..1} = (\frac{n}{p}) (n-1)C(p-1) $

since $ (n-1)C(p-1) $ is a binomial coefficient in its own right, it has to be an integer. Let's call it i.

so $ nCp = \frac{n.i}{p} $

Now here is most important observation of this proof. i is formed by dividing a certain numerator by a certain denominator. The numerator is a product of the terms from (n-1) to (n-p+1). None of these terms are divisible by p. Why? because since n is divisible by a prime p, the next smaller integer that can be divided by p can only be n-p. All the terms of the numerator fall between n and n-p excluding the bounds, so none of them is divisible by p, and neither is their product. So we will call i "p-free".

Since i cannot be divided by p, for $nCp$ to be an int, n has to be divided by p. Since n/p = m, $ nCp = \frac{n.i}{p} = m.i$

Now for n to divide $nCp$, $m.p$ has to divide $m.i$. This gives rise to two options.

m of numerator is divided by m of denominator, and i of numerator is divided by p of denominator.
m of numerator is divided by p of denominator, and i of numerator is divided by m of denominator.

Option 1 is an impossibilty, because it requires i to be divided by p, and we know that i is "p-free".

Option 2 gives rise to two sub-options. Either m contains a factor p, or it does not. If it does not have a factor p, then clearly $ p \nmid m$. And if it has a factor p, then $ m \nmid i $. (remember, i is p-free). So option 2 is also an impossibility.

Once you have that i is "p-free", you are done since you have that i/p is not an integer, that is, nCp/n is not an integer, which is the same as saying n doesn't divide nCp. — Anshuman Agrawal, Mar 03 '24 at 06:15

Proving that $n\mid(nCr)$ for all $r$ ($1 \leq r \leq n-1$), only if $n$ is prime

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