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*}

Introductory

• Let $n$ be a positive integer greater than 4 such that the decimal representation of $n!$ ends in $k$ zeros and the decimal representation of $(2n)!$ ends in $3k$ zeros. Let $s$ denote the sum of the four least possible values of $n$. What is the sum of the digits of $s$? (2015 AMC 10B Problem 23)
• 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)
• Find the sum of digits of the largest positive integer n such that $n!$ ends with exactly $100$ zeros. (demigod)

• 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)