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

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! $\mathrm{}\\$ $\endgroup$ – Ahaan S. Rungta Nov 30 '13 at 0:24

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? $\endgroup$ – Jorge Fernández Hidalgo Dec 25 '13 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. $\endgroup$ – Balarka Sen Dec 25 '13 at 16:55


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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