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On the Wikipedia page about Binomial coefficients, it states that $p$ divides $\binom{p}{k}$ where $p$ is prime. It offers a proof that I don't understand from the very first line. It says that it is true since $\binom{p}{k}$ is a natural number and $p$ divides the numerator but not the denominator. I don't get how that shows us anything really.

Please could you explain why the above statement is true?

Many thanks.

A-Level Student
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    I'm not clear on which part you don't understand. Do you not understand why $\binom pk$ is a natural number? Do you not understand why it's represented by a fraction that has $p$ in the numerator but not in the denominator? Or do you not understand why representing $\binom pk$ as such a fraction yields the conclusion that it must be divisible by $p$? – Robert Shore Mar 25 '21 at 21:44
  • https://math.stackexchange.com/questions/246702/prove-that-p-mid-binompk-0-k-p – Mike Earnest Mar 25 '21 at 21:44
  • @RobertShore sorry for being vague, I'm unclear about the last option. – A-Level Student Mar 25 '21 at 21:45
  • Imagine simplifying $\frac{p(p-1)\cdots(p-k+1)}{k(k-1)\cdots(2)(1)}$ to lowest terms. Since the final result must be an integer, we will cancel out every factor in the denominator with something in the numerator. What is left is a multiple of any factor that remains after cancelling everything (and of any divisor of that factor). But because $p$ is prime, the only way we can cancel out that $p$ in the numerator is with something which is a multiple of $p$ in the denominator. Otherwise, we will never use up that $p$. But nothing in the denominator is a multiple of $p$, so that $p$ "survives." – Arturo Magidin Mar 25 '21 at 21:47
  • @MikeEarnest thank you very much. How can I search the site efficiently so I don't ask duplicate questions? – A-Level Student Mar 25 '21 at 21:47
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    I disagree with the close votes. This is a reasonable question seeking clarity on a specific point of the proof that hasn't been previously highlighted. – Robert Shore Mar 25 '21 at 21:50
  • I found that by searching the keywords "prime divide binomial coefficient," something similar works best for the search bar on MSE. This only works for search words; to search formulas, use https://approach0.xyz/ – Mike Earnest Mar 25 '21 at 21:51
  • @ArturoMagidin thank you for the extremely clear explanation. – A-Level Student Mar 25 '21 at 21:55
  • For the rep you've earned, you continue to ask some very poor questions. – amWhy Mar 25 '21 at 22:16
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    @amWhy I'm sorry you feel that way. Poor in what way? I'd like to know how I can improve them. – A-Level Student Mar 25 '21 at 22:17
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    Of course you disagree, @RobertShore. After all you answered a duplicate question, but still want your credit for giving an answer already given. – amWhy Mar 25 '21 at 22:36
  • @amWhy Really? Which question answered the question the way I did? The other answers I've seen more or less duplicated the Wikipedia article that is linked in the original post, so they clearly were not sufficient to clear up the original poster's confusion. This is exactly the sort of situation where we should be helping, and I submit that the explanation in my answer has not been duplicated elsewhere. It's certain, based on the comments of A-Level Student, that my answer answered the question in a way the poster found useful. – Robert Shore Mar 26 '21 at 01:36

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We know that $\binom pk = n = p \frac ab$ where $a, b$ are integers and $p$ is not a factor of $b$. Therefore, $pa=bn$ so $p \mid bn$. We also know, because $p$ does not divide $b$, that $p$ and $b$ are relatively prime. Therefore, it must be the case that $p$ divides the other factor of the product, which is $n = \binom pk$.

Robert Shore
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