If p is prime number and a and b are natural numbers such that a>b, Prove that

$\large{\dbinom{pa}{pb} \equiv \dbinom{a}{b} \mod p}$

Note by Md Zuhair
2 years, 10 months ago

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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.

You might be interested in the Lucas' theorem wiki.

- 2 years, 10 months ago

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

- 2 years, 10 months ago

U first... hw many?

- 2 years, 10 months ago

@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??

- 2 years, 10 months ago

Did 3. But unsure

- 2 years, 10 months ago

Which ones?

- 2 years, 10 months ago

Inequality, Number theory and Combinatorics

- 2 years, 10 months ago

Can anyone help in this one?

- 2 years, 10 months ago

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

In fact, there is a stronger statement.

For primes $p \geq 5$, ${ pa \choose pb } \equiv { a \choose b } \pmod{p^3}$.

Staff - 2 years, 10 months ago

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

- 2 years, 10 months ago

I doubt that holds for a large range.

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

Staff - 2 years, 10 months ago

For the need of Olympiads only i am learning this.. RMO is this Sunday. :)

Wish a best of luck for me.

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

- 2 years, 10 months ago