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Well, (pbpa)=(pb)!(pa−pb)!(pa)!, and now look at the terms on the top and bottom that are divisible by p. These are
(pb)(p(b−1))(⋯)(p⋅1)(p(a−b))(p(a−b−1))(⋯)(p⋅1)(pa)(p(a−1))(p(a−2))(⋯)(p⋅1)=pbb!⋅pa−b(a−b)!paa!=b!(a−b)!a!=(ba).
What are the leftover terms mod p? It's just a bunch of copies of (p−1)!:
(p−1)!b(p−1)!a−b(p−1)!a≡1.
So that'll do it.
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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
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to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
Well, (pbpa)=(pb)!(pa−pb)!(pa)!, and now look at the terms on the top and bottom that are divisible by p. These are (pb)(p(b−1))(⋯)(p⋅1)(p(a−b))(p(a−b−1))(⋯)(p⋅1)(pa)(p(a−1))(p(a−2))(⋯)(p⋅1)=pbb!⋅pa−b(a−b)!paa!=b!(a−b)!a!=(ba). What are the leftover terms mod p? It's just a bunch of copies of (p−1)!: (p−1)!b(p−1)!a−b(p−1)!a≡1. So that'll do it.
You might be interested in the Lucas' theorem wiki.
@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 (pbpa)≡(ba)modp4 ?
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The p3 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 :)
@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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