Difference between revisions of "Legendre's Formula"
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− | '''Legendre's | + | '''Legendre's Formula''' states that |
− | <cmath>e_p(n)=\sum_{i\ | + | <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>e_p(n)</math> is the exponent of <math>p</math> in the [[prime factorization]] of <math>n!</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== |
− | + | Find the largest integer <math>k</math> for which <math>2^k</math> divides <math>27!</math> | |
+ | ===Solution 1=== | ||
+ | Using the first form of Legendre's Formula, substituting <math>n=27</math> and <math>p=2</math> gives | ||
+ | <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\ | ||
+ | =&13+6+3+1\ | ||
+ | =&23\end{align*}</cmath> | ||
+ | which means that the largest integer <math>k</math> for which <math>2^k</math> divides <math>27!</math> is <math>\boxed{23}</math>. | ||
+ | |||
+ | ===Solution 2=== | ||
+ | Using the second form of Legendre's Formula, substituting <math>n=27</math> and <math>p=2</math> gives | ||
+ | <cmath>e_2(27!)=\frac{27-S_2(27)}{2-1}=27-S_2(27)</cmath> | ||
+ | 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, | ||
+ | <cmath>e_2(27!)=27-S_2(27)=27-4=23</cmath> | ||
+ | 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== | ||
+ | === Part 1 === | ||
+ | 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>\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 === | ||
+ | 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_{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\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*}</cmath> | ||
+ | |||
+ | ===Problems=== | ||
+ | ====Introductory==== | ||
+ | * Let <math>n</math> be a positive integer greater than 4 such that the decimal representation of <math>n!</math> ends in <math>k</math> zeros and the decimal representation of <math>(2n)!</math> ends in <math>3k</math> zeros. Let <math>s</math> denote the sum of the four least possible values of <math>n</math>. What is the sum of the digits of <math>s</math>? (2015 AMC 10B Problem 23) | ||
+ | |||
+ | * How many zeros are at the end of the base-<math>15</math> representation of <math>50!</math>? | ||
+ | |||
+ | * <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) | ||
+ | |||
+ | * Find the sum of digits of the largest positive integer n such that <math>n!</math> ends with exactly <math>100</math> zeros. (demigod) | ||
+ | |||
+ | ====Olympiad==== | ||
+ | * 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 | + | [[Category:Number theory]] |
[[Category:Theorems]] | [[Category:Theorems]] | ||
[[Category:Definition]] | [[Category:Definition]] |
Latest revision as of 19:30, 7 October 2024
Legendre's Formula states that
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 .
Contents
[hide]Examples
Find the largest integer for which divides
Solution 1
Using the first form of Legendre's Formula, substituting and gives which means that the largest integer for which divides is .
Solution 2
Using the second form of Legendre's Formula, substituting and gives The number when expressed in Base-2 is . This gives us . Therefore, which means that the largest integer for which divides is .
Proofs
Part 1
We use a counting argument.
We could say that is equal to the number of multiples of less than , or . But the multiples of are only counted once, when they should be counted twice. So we need to add on. But this only counts the multiples of twice, when we need to count them thrice. Therefore we must add a on. We continue like this to get . This makes sense, because the terms of this series tend to 0.
Part 2
Let the base representation of be where the are digits in base Then, the base representation of is Note that the infinite sum of these numbers (which is ) is
Problems
Introductory
- Let be a positive integer greater than 4 such that the decimal representation of ends in zeros and the decimal representation of ends in zeros. Let denote the sum of the four least possible values of . What is the sum of the digits of ? (2015 AMC 10B Problem 23)
- How many zeros are at the end of the base- representation of ?
- can be written as where and are positive integers. What is the largest possible value of ? (BorealBear)
- Find the sum of digits of the largest positive integer n such that ends with exactly zeros. (demigod)
Olympiad
- Let be numbers of factors of the number (that is, and ). Find the least such that . (Turkey TST 1990)
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