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

(papb)(ab)modp\large{\dbinom{pa}{pb} \equiv \dbinom{a}{b} \mod p}

Note by Md Zuhair
2 years, 2 months ago

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Well, (papb)=(pa)!(pb)!(papb)!,\binom{pa}{pb} = \frac{(pa)!}{(pb)!(pa-pb)!}, and now look at the terms on the top and bottom that are divisible by p.p. These are (pa)(p(a1))(p(a2))()(p1)(pb)(p(b1))()(p1)(p(ab))(p(ab1))()(p1)=paa!pbb!pab(ab)!=a!b!(ab)!=(ab). \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 pp? It's just a bunch of copies of (p1)!(p-1)!: (p1)!a(p1)!b(p1)!ab1. \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.

Patrick Corn - 2 years, 1 month ago

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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?

Md Zuhair - 2 years, 2 months ago

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

In fact, there is a stronger statement.

For primes p5 p \geq 5, (papb)(ab)(modp3) { pa \choose pb } \equiv { a \choose b } \pmod{p^3} .

Calvin Lin Staff - 2 years, 2 months ago

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Sir how do we find the range of p for (papb)(ab)modp4\large{{ pa \choose pb} \equiv {a \choose b} \mod p^4} ?

Md Zuhair - 2 years, 2 months ago

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@Md Zuhair I doubt that holds for a large range.

The p3 p^3 case is also known as wolstenholme's theorem, and the ideas have been used in several olympiad problems.

Calvin Lin Staff - 2 years, 2 months ago

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@Calvin Lin 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 :)

Md Zuhair - 2 years, 2 months ago

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

Aaron Jerry Ninan - 2 years, 1 month ago

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U first... hw many?

Md Zuhair - 2 years, 1 month ago

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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??

Aaron Jerry Ninan - 2 years, 1 month ago

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@Aaron Jerry Ninan Did 3. But unsure

Md Zuhair - 2 years, 1 month ago

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@Md Zuhair Which ones?

Aaron Jerry Ninan - 2 years, 1 month ago

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@Aaron Jerry Ninan Inequality, Number theory and Combinatorics

Md Zuhair - 2 years, 1 month ago

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