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For any value of $N$, is it possible that the factorial of $N$ is equal to a power of 2?

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    $\begingroup$ $2!=2^1$ and $0!=2^0$. Those are the only ones. $\endgroup$ – AugSB Feb 8 at 18:05
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    $\begingroup$ @AugSB also $1!=2^0$ $\endgroup$ – J. W. Tanner Feb 8 at 18:11
  • $\begingroup$ @J.W.Tanner True! I also had that one in mind, but I forgot to add it! $\endgroup$ – AugSB Feb 8 at 18:52
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If $N \ge 3$ then $3$ will divide $N!$ but $3$ will never divide a power of $2$.

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We can find the prime factorization of $N!$ by noting the following:

If we list the prime numbers in order as $p_1, p_2,p_3,.... $ etc. the there is a specific prime $p_n \le N < p_{n+1}$. So the prime factors of $N!$ are $p_1,...., p_n$. A multiple of prime $p_k$ will appear $\lfloor \frac np_{k} \rfloor$ times so $p_k^{\lfloor \frac np_{k} \rfloor}$ will divide $N!$. Furthermore $p_k^2$ will appear $\lfloor \frac n{p_{k}^2} \rfloor$ times and so on.

So $N! = \prod\limits_{p_k\text{ is prime;}\\p_i \le N}p_k^{(\sum\limits^{i=1\\p_k^i\le N}\lfloor \frac n{p_{k}^i} \rfloor)}$

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