How can we find the prime factorization of $100!$. My only idea is to write it as,
$$(1)(2)(3)(\color{red}{2})(\color{red}2)...$$
And count the factors.
But that would take way to long. Any ideas?
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https://en.wikipedia.org/wiki/Legendre's_formula – spaceisdarkgreen Mar 07 '17 at 01:14
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http://math.stackexchange.com/q/573856/ – Jonas Meyer Mar 07 '17 at 01:18
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Hint: use some CAS, by example this. – Masacroso Mar 07 '17 at 01:21
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Here's a start: to determine how many $2$s there are, first consider the even numbers. There are $50$ of those, so that's $50$ factors of $2$. But there are $25$ multiples of $4$, each one adding another factor of $2$; and $12$ multiples of $8$, and $6$ multiples of $16$, $3$ multiples of $32$, and $1$ multiple of $64$. That's a total of $50 + 25 + 12 + 6 + 3 + 1 = 97$ twos. So the prime factorization starts $2^{97} \cdot \ldots$
Reese Johnston
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Thanks it makes sense. But where do I go from here. How do I know which primes to consider? – Ahmed S. Attaalla Mar 07 '17 at 01:24
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@AhmedS.Attaalla The short answer is "all of them". Or at least, all the primes less than $100$. $100!$ can't be divisible by any prime bigger than $100$, so that's as far as you need to go. – Reese Johnston Mar 07 '17 at 01:26