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
9 months, 2 weeks 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.

- 9 months, 1 week ago

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

- 9 months, 1 week ago

U first... hw many?

- 9 months, 1 week 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??

- 9 months, 1 week ago

Did 3. But unsure

- 9 months, 1 week ago

Which ones?

- 9 months, 1 week ago

Inequality, Number theory and Combinatorics

- 9 months, 1 week ago

Can anyone help in this one?

- 9 months, 2 weeks 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 - 9 months, 2 weeks ago

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

- 9 months, 2 weeks 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 - 9 months, 2 weeks 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 :)

- 9 months, 2 weeks ago