Well, $\binom{pa}{pb} = \frac{(pa)!}{(pb)!(pa-pb)!},$ and now look at the terms on the top and bottom that are divisible by $p.$ These are
$\frac{(pa)(p(a-1))(p(a-2))(\cdots)(p\cdot 1)}{(pb)(p(b-1))(\cdots)(p\cdot 1)(p(a-b))(p(a-b-1))(\cdots)(p\cdot 1)} = \frac{p^a a!}{p^b b! \cdot p^{a-b} (a-b)!} = \frac{a!}{b!(a-b)!} = \binom{a}{b}.$
What are the leftover terms mod $p$? It's just a bunch of copies of $(p-1)!$:
$\frac{(p-1)!^a}{(p-1)!^b(p-1)!^{a-b}} \equiv 1.$
So that'll do it.

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## Comments

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TopNewestWell, $\binom{pa}{pb} = \frac{(pa)!}{(pb)!(pa-pb)!},$ and now look at the terms on the top and bottom that are divisible by $p.$ These are $\frac{(pa)(p(a-1))(p(a-2))(\cdots)(p\cdot 1)}{(pb)(p(b-1))(\cdots)(p\cdot 1)(p(a-b))(p(a-b-1))(\cdots)(p\cdot 1)} = \frac{p^a a!}{p^b b! \cdot p^{a-b} (a-b)!} = \frac{a!}{b!(a-b)!} = \binom{a}{b}.$ What are the leftover terms mod $p$? It's just a bunch of copies of $(p-1)!$: $\frac{(p-1)!^a}{(p-1)!^b(p-1)!^{a-b}} \equiv 1.$ So that'll do it.

You might be interested in the Lucas' theorem wiki.

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@Calvin Lin @Brilliant Mathematics @Chew-Seong Cheong @Pi Han Goh @Kushal Bose @Aaron Jerry Ninan

Can anyone help in this one?

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This is a lemma in the proof of lucas' theorem.

In fact, there is a stronger statement.

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Sir how do we find the range of p for $\large{{ pa \choose pb} \equiv {a \choose b} \mod p^4}$ ?

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The $p^3$ case is also known as wolstenholme's theorem, and the ideas have been used in several olympiad problems.

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Wish a best of luck for me.

You were one of those people who represented the your motherland. Proud of you :)

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@Md Zuhair @Rahil Sehgal How may questions did you solve in RMO 2017 ???

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U first... hw many?

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@Md Zuhair I solved 4, but I didnt get the time to write 1 question completely....so effectively I did 3.5... What abt you??

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