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I know that $100!$ is $100\times99\times98\times...2\times1$ but I don't get how to get all the factors of three from this. Help!

Vidyanshu Mishra
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jujibaby
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    How many multiples of three are there less than 100? How many multiples of $9$? How many multiples of $27$? How many multiples of $81$? – JMoravitz Nov 30 '16 at 18:54
  • Start small. How many factors of $3$ in $10!$ what is the pattern? Can you prove that the pattern holds as the numbers get bigger? – Doug M Nov 30 '16 at 19:00
  • @JMoravitz I don't think so. The process is the same but the question is a different one, that question reduces to figuring out the 5-adic valuation of 100! whereas this one is the 3-adic valuation – operatorerror Nov 30 '16 at 19:13
  • @qbert the theorem cited in the answer applies directly to this question as well – JMoravitz Nov 30 '16 at 19:14
  • @JMoravitz that may be, but we don't close every question that uses the mean value theorem – operatorerror Nov 30 '16 at 19:14

5 Answers5

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Hints:

  • Every multiple of 3 introduces a factor 3 (like 30, 96), how many of those numbers are there in 100!
  • Every multiple of 9 introduces an additional factor 3 (like 36, 99), how many?
  • Every multiple of 27 yet another one, and every factor 81 even one more.
Pieter21
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You are looking for the 3-adic valuation of 100!, which is given by the Legendre formula $$ \lfloor \frac{100}{3} \rfloor + \lfloor \frac{100}{3^2} \rfloor+....\\ =33+11+3+1=48 $$

operatorerror
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The only thing you have to do here is to calculate the highest power of $3$ contained in $100!$. It can be easily calculated using de polignac's formula.

For your question it looks like $[\frac{100}{3}]+[\frac{100}{9}]+[\frac{100}{27}]+[\frac{100}{81}]=48$

Vidyanshu Mishra
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$$\color\red{\sum\limits_{n=1}^{\lfloor\log_{3}{100}\rfloor}\left\lfloor\frac{100}{3^n}\right\rfloor}=\left\lfloor\frac{100}{3}\right\rfloor+\left\lfloor\frac{100}{9}\right\rfloor+\left\lfloor\frac{100}{27}\right\rfloor+\left\lfloor\frac{100}{81}\right\rfloor=33+11+3+1=48$$

barak manos
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Here's a not conventional way to think of it but... it might help:

$100! = \prod_{k=1}^{100}k$

$=100*\prod_{k=1}^{99}k$

$=100*\prod_{k=1}^{33}3k*\prod_{j=1}^{33}(3j-1)*\prod_{l=1}^{33}(3l-2)$

$3\not \mid 100*\prod_{j=1}^{33}(3j+-1)\prod_{l=1}^{33}(3l-2)$

so $100!$ and $\prod_{k=1}^{33}3k$ have the same power of $3$ factors.

$\prod_{k=1}^{33}3k=3^{33}\prod_{k=1}k$

$= 3^{33}*\prod_{k=1}^{11}3k*\prod_{j=1}^{11}(3j-1)\prod_{l=1}^{11}(3l-2)$

$3\not \mid \prod_{j=1}^{11}(3j-1)\prod_{l=1}^{11}(3l-2)$

so $100!$ and $3^{33}\prod_{k=1}^{10}3k$ have the same power of $3$ factors.

$3^{33}\prod_{k=1}^{11}3k = 3^{33}3^{11}\prod_{k=1}^{11}k$

$=3^{33}3^{11}*11*10*\prod_{k=1}^33k*\prod_{j=1}^3(3j-1)*\prod_{l=1}^3(3j-2)$

And again $3 \not \mid 10*11*\prod_{j=1}^3(3j-1)*\prod_{l=1}^3(3j-2)$ so

$100!$ has same power of $3$ factor of $3^{33}3^{11}\prod_{k=1}^33k$

$3^{33}3^{11}\prod_{k=1}^33k= 3^{33}3^{11}3^{3}\prod_{k=1}^3 k$

$3^{33}3^{11}3^{3}*1*2*3$ and $3 \not \mid 2$ so

$3^{33}3^{11}3^{3}*3 = 3^{33+11+3+1} = 3^{48}$ is the highest power of $3$ that divide $100!$.

In other words $\lfloor 100/3 \rfloor=33$ of the integer terms of $1*2*3....*98*99*100$ are multiples of $3$ so $3^{33}|100!$. Of those $33$ multiples of $3$ one third of them are multiples of $9$ and so on.

fleablood
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