Difference between revisions of "Legendre's Formula"
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<cmath>e_p(n)=\sum_{i\geq 1} \left\lfloor \dfrac{n}{p^i}\right\rfloor =\frac{n-S_{p}(n)}{p-1}</cmath> | <cmath>e_p(n)=\sum_{i\geq 1} \left\lfloor \dfrac{n}{p^i}\right\rfloor =\frac{n-S_{p}(n)}{p-1}</cmath> | ||
| − | where <math>e_p(n)</math> is the [[exponent]] of <math>p</math> in the [[prime factorization]] of <math>n!</math> and <math>S_p(n)</math> is the [[sum]] of the [[digit]]s of <math>n</math> when written in [[base]] <math>p</math>. | + | where <math>p</math> is a prime and <math>e_p(n)</math> is the [[exponent]] of <math>p</math> in the [[prime factorization]] of <math>n!</math> and <math>S_p(n)</math> is the [[sum]] of the [[digit]]s of <math>n</math> when written in [[base]] <math>p</math>. |
==Proof== | ==Proof== | ||
Revision as of 06:53, 15 April 2009
Legendre's Formula states that
![\[e_p(n)=\sum_{i\geq 1} \left\lfloor \dfrac{n}{p^i}\right\rfloor =\frac{n-S_{p}(n)}{p-1}\]](http://latex.artofproblemsolving.com/7/a/e/7aec74a5d87d77aaead655f9dc973202626f8093.png)
where 
is a prime and 
is the exponent of in the prime factorization of 
and 
is the sum of the digits of 
when written in base .
Proof
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