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Using only three 1s and two 2s (i.e. 1,1,1,2,2), can you generate the number 55? You can use all mathematical operators you know, such as floor function, factorial, powers/indices*, etc., but concatenation is not allowed (e.g. '11' from 1 and 1 is not permitted).

* Any powers used must be derived from the initial set of three 1s and two 2s - you don't get to use 'squared' (for example) without accounting for the '2' required when writing this as a formula.

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    $\begingroup$ 11*(1+2+2) feels natural $\endgroup$
    – Daniel S
    Oct 17, 2023 at 11:33
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    $\begingroup$ @Daniel "example, 11 from 1,1 is not allowed". $\endgroup$ Oct 17, 2023 at 12:22
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    $\begingroup$ Is the "decimal point magic" mentioned below legit? I feel like using .2 brings in an extra number, 10. Also, what about the exp function, which brings in e? $\endgroup$ Oct 17, 2023 at 23:05
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    $\begingroup$ My intuition says that using floor or ceiling (even if it's explicitly allowed in the question) should be less elegant than "moderately esoteric" approaches like tetration or double factorial; it can do too much. Out of the approaches seen so far, I feel like binomial coefficients are the cleanest, and "decimal point magic" the second cleanest (because of the concern that $.2$ is really just short for the forbidden $0.2$, and also perilously close to concatenation). $\endgroup$ Oct 18, 2023 at 1:09
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    $\begingroup$ Hey guys I am surprised no one mentioned this, but isn't this just straight copied from this question on Math SE? Checking the timestamps indicate that this question has been posted about 3 hours after the one on Math SE was posted. $\endgroup$
    – Aiden Chow
    Oct 20, 2023 at 6:50

13 Answers 13

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My answer is

$ 55 = 1 + (1 + 2)! + ((1 + 2)!)!! $

$ 55 = 1 + 3! + (3!)!! $
$ 55 = 1 + 6 + 6!! $
$ 55 = 7 + 48 $


Another solution, again with no concatentation

$ 55 = \sqrt{(((1 + 2)!)!! + 1)} + ((1 + 2)!)!! $

$ 55 = \sqrt{((3!)!! + 1)} + (3!)!! $
$ 55 = \sqrt{(6!! + 1)} + 6!! $
$ 55 = \sqrt{(48 + 1)} + 48 $
$ 55 = 7 + 48 $


A solution using base 6, working in the other direction.

$ 55 = 100 - 1 $
$ 55 = 10^2 - 1 $
$ 55 = 3!^2 - 1 $

$ 55 = (1 + 2)!^2 - 1^1 $

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    $\begingroup$ +1 for using base six. Lovely little loophole that $\endgroup$
    – No Name
    Oct 18, 2023 at 0:56
  • $\begingroup$ Aw shucks, I was going for the double factorial as well (+1) $\endgroup$ Oct 20, 2023 at 8:57
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Using only addition, division, and decimal point magic, while preserving the order of the digits:

$$\frac{1+\frac{1+1}{.2}}{.2}=\frac{11}{.2}=55$$

Or alternately, using some sillier operators and a nifty coincidence:

$$ \sum_{n=1}^{(1+2)!} (n-1)^2 $$ $$ = 0 + 1 + 4 + 9 + 16 + 25 = 55 $$

Or utilising combinatorics:

$$ 1/.1 - 1 + 2 \choose{2} $$ $$= 11 \text{ choose } 2 = 10+9+8+7+6+5+4+3+2+1 = 55$$

Or finally (always save the prettiest one for last), bringing in the notation $T_n$ for the nth triangular number:

$$T_{T_{1 \times 1 \times 1 \times 2 \times 2}} $$ $$=T_{T_4} = T_{10} = 55$$

which can (if you don't mind the slightly awkward MathJax layout) also be written using exponents for some further savings on ink (and/or pixels):

$$T_{T_{2^{2^{1^{1^{1}}}}}}$$

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    $\begingroup$ Decimal point seems a bit magic. You're just bringing a 10 to the table. The ones NOT using decimal point are very clever. $\endgroup$ Oct 19, 2023 at 10:09
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    $\begingroup$ If you're not allowed to make 11 from 1 and 1 I don't think you should be allowed to make .1 from 1. But +1 for the first and last examples, they're great. $\endgroup$
    – N. Virgo
    Oct 20, 2023 at 7:03
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Since "indices" are allowed, how about we

index into the Fibonacci numbers, with $F_{(1+1)(1+2+2)}$ giving us the $10^{\text{th}}$ Fibonacci number, $55$?

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    $\begingroup$ This one's a good one $\endgroup$
    – MWQOJYNWQA
    Oct 17, 2023 at 20:40
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    $\begingroup$ I have heard rumors that some places, "indices" is mostly another name for exponents. Which makes sense given the context in which they are mentioned in the OP. $\endgroup$
    – Arthur
    Oct 18, 2023 at 12:21
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Ends up with an approximation, but a handy ceiling function makes it legitimate.

$\lceil(\sqrt{1+1}+(1+2)!)^2\rceil$

⸂⸂⸜(രᴗര๑)⸝⸃⸃ woohoo

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$-\log_2(\log_2(\sqrt(\sqrt(\sqrt(...\sqrt(1 + 1 * 1) ... )))))$

with 55 total square root signs.

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  • $\begingroup$ Sneaky way to get all the 1/2's in. $\endgroup$ Oct 20, 2023 at 7:50
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    $\begingroup$ +1 for an answer that generalises to other target values $\endgroup$ Oct 20, 2023 at 14:25
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As a binomial coefficient

$$\begin{pmatrix}\sqrt {((2+1)!-1)!+1} \\ 2 \end{pmatrix}$$.

If we accept Donald Knuth's notation n? for the n-th triangular number this can also be written

$$(1+2?)??$$

with a 2 and two 1s to spare.

Alternatively, here is one inspired by @Lynn's tetration:

$$\sqrt{\sqrt{.2}^{-.1^{-1}}-.1^{-2}} = \sqrt{\sqrt 5^{10}-10^2} = \sqrt{3125-100} = 55$$

Last not least, as absolute value of a complex number:

$$\left | ((2+1)!)!! + \sqrt{-((2+1)!)!-1} \right| = \left | 48 + \sqrt{721}\mathrm i \right | = \sqrt{48^2+721} = 55$$

If we use triangular numbers as above we can write more compactly:

$$\left | 2??!! + \sqrt{-2??!-1} \right|$$

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Using one tetration:

$$\left\lfloor \sqrt{^2 (1+1+1+2)} \right\rfloor = \lfloor \sqrt{5^5} \rfloor = \lfloor \sqrt{3125} \rfloor = \lfloor 55.901\dots \rfloor = 55.$$

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I tried using as few numbers as possible, which gave

$ (\frac{2}{.2} + 1)!!!!!! $
$ = (10 + 1)!!!!!! $
$ = 11!!!!!! $
$ = 11 \times 5 $
$ = 55 $

We can also easily add the additional numbers to this with

$ (\frac{2}{.2} + 1)!!!!!! + 1 - 1$

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  • $\begingroup$ How is the operator '!' defined here? It does not appear to be the common definition of the factorial. $\endgroup$
    – xyldke
    Oct 18, 2023 at 10:00
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    $\begingroup$ @xyldke the multifactorial is essentially a factorial that skips numbers. For instance, the double factorial n!! = n(n-2)(n-4)... Here, the 6-th multifactorial would be n(n-6) = 11*5 $\endgroup$
    – Bryce
    Oct 18, 2023 at 14:19
  • $\begingroup$ I don't want to edit my post after the fact but I just realized $((2+2)!!)_{1+1}-1$ also works using both multifactorial & falling factorials and $(2+2)??$ works using triangular numbers since $T_4 = 10$ and $T_10 = 55$! $\endgroup$
    – Bryce
    Oct 22, 2023 at 17:49
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Using only two of the digits!

$\lceil \exp(2+2) \rceil = \lceil 54.59815... \rceil = 55 $

And you can consume the ones by doing
$ \times 1 \times 1 \times 1 $

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    $\begingroup$ I posted a similar answer before I realized you beat me to it! I would argue that exp counts since this can be written as $\lceil\lim_{n \to \infty} (1 + \frac{2 + 2}{n})^n\rceil$ which still uses only one extra 1 in addition to the two 2s $\endgroup$ Oct 19, 2023 at 3:17
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$55 = 1 + \large \Bigl( \sqrt \frac{1}{.\overline{1}} \Bigr)! \times \frac{2}{.\overline{2}}$

where $.\overline{n}$ means $.nnnnn\cdots$

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  • $\begingroup$ You need to apply floor to sqrt, otherwise you'd be using the non-integer generalization of factorial $\endgroup$
    – smci
    Oct 18, 2023 at 1:26
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    $\begingroup$ @smci No, the repeating decimal makes it $\sqrt{\frac{1}{1/9}}=3$. $\endgroup$ Oct 18, 2023 at 9:05
  • $\begingroup$ Ah, my mistake. $\endgroup$
    – smci
    Oct 18, 2023 at 21:35
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Inspired by ralphmerridew's answer, using only one digit, you can make any number:

$\lceil-\ln(\ln(\sqrt{\sqrt{\sqrt{...\sqrt{2}}}}))\rceil$

Where every square root increases the total by $\ln(2) = 0.693$

To make 55, use 78 sqrts.


Other solutions:

Using only 4 of the digits:

$((1/.1)+1)/.2$

$= (10+1)/.2$
$= 11/.2$
$= 55$

If we want to use all 5, just change it slightly to this:

$(((2-1)/.1)+1)/.2$

Using 4 digits, without using decimal points:

$\lceil\sqrt{(((2+1)!)!)}\rceil * 2 + 1$

$= \lceil\sqrt{6!}\rceil * 2 + 1$
$= \lceil\sqrt{720}\rceil * 2 + 1$
$= \lceil 26.8 \rceil * 2 + 1$
$= 27 * 2 + 1 = 55$

Or with 5 digits like this:

$\lceil\sqrt{1+(((2+1)!)!)}\rceil * 2 + 1$

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Using the generalized multifactorial, we can get the following

$$((2+1)! + (2+1)! - 1)!!!!!! = 11 \times 5 = 55$$

Here's a slightly more convoluted answer using the floor function, the cotangent function and binomial coefficient

$$\binom{-\lfloor \cot(2+1) \rfloor}{2+1} - 1 = 56 -1 = 55$$

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-5
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$(1+1)^3 - 1 + 2^4 + 2^5$

$= 2^3 - 1 + 2^4 + 2^5$

$= 8 - 1 + 16 + 32$

$= 7 + 16 + 32$

$= 7 + 48$

= 55

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    $\begingroup$ Here you're using a '3', a '4' and a '5' as powers without having accounted for them - these would need to be calculated using the original 1,1,1,2,2 as per the question specification. $\endgroup$
    – Stiv
    Oct 17, 2023 at 12:31

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