Difference between revisions of "Legendre's Formula"

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Legendre's Formula states that

$e_p(n!)=\sum_{i=1}^{\infty} \left\lfloor \dfrac{n}{p^i}\right\rfloor =\frac{n-S_{p}(n)}{p-1}$

where $p$ is a prime and $e_p(n!)$ is the exponent of $p$ in the prime factorization of $n!$ and $S_p(n)$ is the sum of the digits of $n$ when written in base $p$ .

Examples

Find the largest integer $k$ for which $2^k$ divides $27!$

Solution 1

Using the first form of Legendre's Formula, substituting $n=27$ and $p=2$ gives $\begin{align*}e_2(27!)=&\left\lfloor\frac{27}{2}\right\rfloor+\left\lfloor\frac{27}{2^2}\right\rfloor+\left\lfloor\frac {27}{2^3}\right\rfloor+\left\lfloor\frac{27}{2^4}\right\rfloor\\ =&13+6+3+1\\ =&23\end{align*}$ which means that the largest integer $k$ for which $2^k$ divides $27!$ is $\boxed{23}$ .

Solution 2

Using the second form of Legendre's Formula, substituting $n=27$ and $p=2$ gives $e_2(27!)=\frac{27-S_2(27)}{2-1}=27-S_2(27)$ The number $27$ when expressed in Base-2 is $11011$ . This gives us $S_2(27)=1+1+0+1+1=4$ . Therefore, $e_2(27!)=27-S_2(27)=27-4=23$ which means that the largest integer $k$ for which $2^k$ divides $27!$ is $\boxed{23}$ .

Proofs

Part 1

We use a counting argument.

We could say that $e_p(n!)$ is equal to the number of multiples of $p$ less than $n$ , or $\left\lfloor \frac{n}{p}\right\rfloor$ . But the multiples of $p^2$ are only counted once, when they should be counted twice. So we need to add $\left\lfloor \frac{n}{p^2}\right\rfloor$ on. But this only counts the multiples of $p^3$ twice, when we need to count them thrice. Therefore we must add a $\left\lfloor \frac{n}{p^3}\right\rfloor$ on. We continue like this to get $e_p(n!)=\sum_{i=1}^{\infty} \left\lfloor \dfrac{n}{p^i}\right\rfloor$ . This makes sense, because the terms of this series tend to 0.

Part 2

Let the base $p$ representation of $n$ be $e_xe_{x-1}e_{x-2}\dots e_0$ where the $e_i$ are digits in base $p.$ Then, the base $p$ representation of $\left\lfloor \frac{n}{p^i}\right\rfloor$ is $e_xe_{x-1}\dots e_{x-i}.$ Note that the infinite sum of these numbers (which is $e_p(n!)$ ) is

$\begin{align*} \sum_{j=1}^{x} e_j\cdot(p^{j-1}+p^{j-2}+\cdots +1) &= \sum_{j=1}^{x} e_j \left( \frac{p^j-1}{p-1} \right) \\ &=\frac{\sum_{j=1}^{x} e_jp^j -\sum_{j=1}^{x} e_j}{p-1} \\ &=\frac{(n-e_0)-(S_p(n)-e_0)}{p-1} \\ &=\frac{n-S_p(n)}{p-1}. \end{align*}$

Problems

Introductory

Olympiad

Let $b_m$ be numbers of factors $2$ of the number $m!$ (that is, $2^{b_m}|m!$ and $2^{b_m+1}\nmid m!$ ). Find the least $m$ such that $m-b_m = 1990$ . (Turkey TST 1990)

This article is a stub. Help us out by expanding it.

@@ Line 3: / Line 3: @@
 <cmath>e_p(n!)=\sum_{i=1}^{\infty} \left\lfloor \dfrac{n}{p^i}\right\rfloor =\frac{n-S_{p}(n)}{p-1}</cmath>
-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>.
+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>.
 ==Examples==

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