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So I say this puzzle online a few days ago and found it quite interesting. The original question was

Make $120$ using only five $0$s.

Well, I said to myself, this is utterly trivial. Note that $$ 120 = 5! = (0! + 0! + 0! + 0! + 0!)!. $$ But what if we want to do it for an arbitrary number $n$ and an arbitrary number of $0$s, $m$. That is: we want to make $n$ using only $m$ zeroes. Clearly, using my solution above, we can make $n=m!$ using $m$ zeroes.

For $n=121$ and $m=5$, this is tougher and I can't seem to find a solution. Does anybody want to try to take on some general cases?

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    $\begingroup$ $120 = ((0!+0!+0!)!-0!)!$, only four zeros. $\endgroup$ Commented Nov 28, 2013 at 16:13
  • $\begingroup$ Out of curiosity, why was this downvoted? Also, how come the person who favorited this did not upvote it? $\endgroup$ Commented Nov 28, 2013 at 16:13
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    $\begingroup$ Fischer was answering the $121$ problem, explaining that we can do $$ 121 = ((0! + 0! + 0!)! - 0!)! + 0! $$ $\endgroup$ Commented Nov 28, 2013 at 16:18
  • $\begingroup$ @AkshajKadaveru Nice to see you here! $\endgroup$ Commented Nov 28, 2013 at 16:19

3 Answers 3

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I'm not sure if this type of solution is what you're looking for, but this sort of problem is pretty trivial if you don't restrict the set of allowed operators somehow:

$$ 121 = \tan \arccos \underbrace{\sin \arctan \sin \arctan \cdots \sin \arctan}_{121^2 \textrm{ copies of}\sin \arctan} \cos 0$$

See USAMO 1995.2.

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    $\begingroup$ Brilliant! $\mathrm{}\\$ $\endgroup$ Commented Nov 30, 2013 at 0:24
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A Perelman-like solution might suffice :

$$121 = -\log_2 \log_2 \underbrace{\sqrt{\sqrt{\sqrt{\cdots\sqrt{(0!+0!)!}}}}}_\text{121 copies square roots}$$

Which uses only two $0$'s, next best to betaveros. $$$$

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  • $\begingroup$ why do you say it is a perelman like solution? $\endgroup$
    – Asinomás
    Commented Dec 25, 2013 at 16:51
  • $\begingroup$ Most people would think I am referring to Gregori Perelman, which is not the case. This is referred to physicist Yakov Perelman, and something alike is described in one of his publications, where he says this was first shown in a congress of physicist in Odessa. I couldn't find further reference about that congress, so I eventually named it after him. $\endgroup$ Commented Dec 25, 2013 at 16:55
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Note:

If we could define the quadruple factorial formula with some symbols be can get $120$ by using only three $0$'s, because quadruple factorial is formed by formula $\frac{(2n)!}{n!}$.

Reference link here

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