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Lucas's theorem

Keywords: 
binomial coefficient,congruence
Type of Math Object: 
Theorem
Major Section: 
Reference

Mathematics Subject Classification

11B65 no label found

Comments

Hi, Slash,
think you were not quite precise in your statement. Consider this:
Let $m,n$ be natural numbers, $m \ge n$ and $p$ a prime number dividing both
$m$ and $n$. Then you have
\[{m \choose n} =\frac{m^{'}.(m -1)\cdots (m -(n -1))}{p^s.n^{'}(n -1)\cdots
1}\]
where $r,s$ are the maximum powers of $p$ dividing $m$ and $n$. But then the
binomial coefficient would be $0 \bmod p$, but ${0 \choose 0} =1$ so in this
case the theorem is wrong.

I'm sure I found this theorem in a paper about binomial coefficients,
written by Andrew Granville. This paper sure also contains a proof of
this...

Regards,

Thomas

If r>s , then the number of zeros at the end in the p-base expansion of m is greater than the one in the expansion of n , so you would get that the product is zero ! (a term of the form {0 \choose something} appears!) So .. the theorem is true in this case (correct me if I'm wrong)! I have a proof of it but I didn't have time to write it up in LaTeX .

Cheers !

P.S. : I hope I'm not talking nonsense and that it's correct

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