I want to prove that:
$2^{n-1}\mid n!$ if and only if $n$ is a power of $2$.
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Hint: It is well-known(?) that the maximal $k$ with $p^k|n!$ is given by $$k=\left\lfloor \frac{n}{p}\right\rfloor+\left\lfloor \frac{n}{p^2}\right\rfloor+\left\lfloor \frac{n}{p^3}\right\rfloor+\ldots$$
Hagen von Eitzen
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this formula is unknown for me? and then? – misi10 Nov 30 '12 at 15:44
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This is sometimes known as De Polignac's formula but is also attributed to Legendre. – lhf Nov 30 '12 at 16:02
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i can not continue,give me more hint. – misi10 Nov 30 '12 at 18:47
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1Hagen, http://math.stackexchange.com/questions/141196/highest-power-of-a-prime-p-dividing-n and many other MSE questions. – Will Jagy Nov 30 '12 at 19:56
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1There is also some recent thing for posting fixed, easily searchable quality answers to questions, or parts of questions, that come up over and over and over again. I re-posted my little induction proof there but i do not recall how to find it. Certainly, if some fact is necessary for 100 different questions (this probably qualifies), a post with a few correct proofs seems worthwhile. – Will Jagy Nov 30 '12 at 20:05
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https://en.wikipedia.org/wiki/Legendre%27s_formula – user236182 Mar 11 '16 at 09:46
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write out n! and consider how many times each divides by 2 for example 16:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 = 16!
^ ^ ^ ^ ^ ^ ^ ^ divisible by 2
^ ^ ^ ^ divisible by 2^2
^ ^ divisible by 2^3
^ divisible by 2^4
so there are $16/2 = 8$ numbers there divisible by 2. $16/4 = 4$ numbers there divisible by 2^2. etc.
in total then, there are 16/2 + 16/2^2 + 16/2^3 + 16/2^4.
In general if 2^r is the highest power of 2 dividing n! then $r = \sum_{i=1} [n/2^i]$.
To show that $2^{n-1}\mid n!$ we thus want to prove $n-1 = \sum_{i=1} [n/2^i]$ iff $n = 2^k$. ($\Leftarrow$) is immediate from computation. For ($\Rightarrow$) Let $2^{k-1} < n < 2^k$ but then $r < n - 1$ by comparing each term of the summation for $n$ and the one for $2^{k}$.