Difference between revisions of "Legendre's Formula"

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<cmath>e_p(n!)=\sum_{i=1}^{\infty} \left\lfloor \dfrac{n}{p^i}\right\rfloor =\frac{n-S_{p}(n)}{p-1}</cmath>
 
<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>.
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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>.
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 +
==Examples==
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Find the largest integer <math>k</math> for which <math>2^k</math> divides <math>27!</math>
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 +
===Solution 1===
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Using the first form of Legendre's Formula, substituting <math>n=27</math> and <math>p=2</math> gives
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<cmath>\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\\
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=&13+6+3+1\\
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=&23\end{align*}</cmath>
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which means that the largest integer <math>k</math> for which <math>2^k</math> divides <math>27!</math> is <math>\boxed{23}</math>.
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===Solution 2===
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Using the second form of Legendre's Formula, substituting <math>n=27</math> and <math>p=2</math> gives
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<cmath>e_2(27!)=\frac{27-S_2(27)}{2-1}=27-S_2(27)</cmath>
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The number <math>27</math> when expressed in Base-2 is <math>11011</math>. This gives us <math>S_2(27)=1+1+0+1+1=4</math>. Therefore,
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<cmath>e_2(27!)=27-S_2(27)=27-4=23</cmath>
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which means that the largest integer <math>k</math> for which <math>2^k</math> divides <math>27!</math> is <math>\boxed{23}</math>.
  
 
==Proofs==
 
==Proofs==
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We use a counting argument.
 
We use a counting argument.
  
We could say that <math>e_p(n!)</math> is equal to the number of multiples of <math>p</math> less than <math>n</math>, or <math>\lfloor \frac{n}{p}\rfloor</math>. But the multiples of <math>p^2</math> are only counted once, when they should be counted twice. So we need to add <math>\lfloor \frac{n}{p^2}\rfloor</math> on. But this only counts the multiples of <math>p^3</math> twice, when we need to count them thrice. Therefore we must add a <math>\lfloor \frac{n}{p^3}\rfloor</math> on. We continue like this to get <math>e_p(n!)=\sum_{i=1}^{\infty} \left\lfloor \dfrac{n}{p^i}\right\rfloor</math>. This makes sense, because the terms of this series tend to 0.
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We could say that <math>e_p(n!)</math> is equal to the number of multiples of <math>p</math> less than <math>n</math>, or <math>\left\lfloor \frac{n}{p}\right\rfloor</math>. But the multiples of <math>p^2</math> are only counted once, when they should be counted twice. So we need to add <math>\left\lfloor \frac{n}{p^2}\right\rfloor</math> on. But this only counts the multiples of <math>p^3</math> twice, when we need to count them thrice. Therefore we must add a <math>\left\lfloor \frac{n}{p^3}\right\rfloor</math> on. We continue like this to get <math>e_p(n!)=\sum_{i=1}^{\infty} \left\lfloor \dfrac{n}{p^i}\right\rfloor</math>. This makes sense, because the terms of this series tend to 0.
  
 
=== Part 2 ===
 
=== Part 2 ===
Let the base <math>p</math> representation of <math>n</math> be <cmath>e_xe_{x-1}e_{x-2}\dots e_0</cmath> where the <math>e_i</math> are digits in base <math>p.</math> Then, the base <math>p</math> representation of <math>\lfloor \frac{n}{p^i}\rfloor</math> is <cmath>e_xe_{x-1}\dots e_{x-i}.</cmath> Note that the infinite sum of these numbers (which is <math>e_p(n!)</math>) is
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Let the base <math>p</math> representation of <math>n</math> be <cmath>e_xe_{x-1}e_{x-2}\dots e_0</cmath> where the <math>e_i</math> are digits in base <math>p.</math> Then, the base <math>p</math> representation of <math>\left\lfloor \frac{n}{p^i}\right\rfloor</math> is <cmath>e_xe_{x-1}\dots e_{x-i}.</cmath> Note that the infinite sum of these numbers (which is <math>e_p(n!)</math>) is
  
<cmath>\begin{align*} \sum_{j=1}^{x} e_j(p^{j-1}+p^{j-2}+\cdots +1) &= \sum_{j=1}^{x} e_j \left( \frac{p^j-1}{p-1} \right) \\
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<cmath>\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{\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-e_0)-(S_p(n)-e_0)}{p-1} \\
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\end{align*}</cmath>
 
\end{align*}</cmath>
  
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===Problems===
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==== Introductory====
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* How many zeros are at the end of the base-<math>15</math> representation of <math>50!</math>?
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* <math> \binom{4042}{2021}+\binom{4043}{2022} </math> can be written as <math> n\cdot 10^x </math> where <math> n </math> and <math> x </math> are positive integers. What is the largest possible value of <math> x </math>? (BorealBear)
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====Olympiad====
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* Let <math>b_m</math> be numbers of factors <math>2</math> of the number <math>m!</math> (that is, <math>2^{b_m}|m!</math> and <math>2^{b_m+1}\nmid m!</math>). Find the least <math>m</math> such that <math>m-b_m = 1990</math>. (Turkey TST 1990)
 
{{stub}}
 
{{stub}}
 
[[Category:Number theory]]
 
[[Category:Number theory]]
 
[[Category:Theorems]]
 
[[Category:Theorems]]
 
[[Category:Definition]]
 
[[Category:Definition]]

Latest revision as of 20:05, 27 April 2021

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

  • How many zeros are at the end of the base-$15$ representation of $50!$?
  • $\binom{4042}{2021}+\binom{4043}{2022}$ can be written as $n\cdot 10^x$ where $n$ and $x$ are positive integers. What is the largest possible value of $x$? (BorealBear)

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)

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