# Let's make a huge number with only tiny numbers

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.

• 11*(1+2+2) feels natural Oct 17 at 11:33
• @Daniel "example, 11 from 1,1 is not allowed". Oct 17 at 12:22
• 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? Oct 17 at 23:05
• 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). Oct 18 at 1:09
• 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. Oct 20 at 6:50

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

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

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.

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

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

• +1 for using base six. Lovely little loophole that Oct 18 at 0:56
• Aw shucks, I was going for the double factorial as well (+1) Oct 20 at 8:57

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):

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

• Decimal point seems a bit magic. You're just bringing a 10 to the table. The ones NOT using decimal point are very clever. Oct 19 at 10:09
• 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. Oct 20 at 7:03

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, $$55$$?

• This one's a good one Oct 17 at 20:40
• 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. Oct 18 at 12:21

Ends up with an approximation, but a handy ceiling function makes it legitimate.

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

⸂⸂⸜(രᴗര๑)⸝⸃⸃ woohoo

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

with 55 total square root signs.

• Sneaky way to get all the 1/2's in. Oct 20 at 7:50
• +1 for an answer that generalises to other target values Oct 20 at 14:25

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:

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

Using one tetration:

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

I tried using as few numbers as possible, which gave

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

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

• How is the operator '!' defined here? It does not appear to be the common definition of the factorial. Oct 18 at 10:00
• @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 Oct 18 at 14:19
• 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 $T_10 = 55$! Oct 22 at 17:49

Using only two of the digits!

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

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

• 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 Oct 19 at 3:17

$$55 = 1 + \large \Bigl( \sqrt \frac{1}{.\overline{1}} \Bigr)! \times \frac{2}{.\overline{2}}$$

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

• You need to apply floor to sqrt, otherwise you'd be using the non-integer generalization of factorial
– smci
Oct 18 at 1:26
• @smci No, the repeating decimal makes it $\sqrt{\frac{1}{1/9}}=3$. Oct 18 at 9:05
• Ah, my mistake.
– smci
Oct 18 at 21:35

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:

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

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

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

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

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

$$= 8 - 1 + 16 + 32$$

$$= 7 + 16 + 32$$

$$= 7 + 48$$

= 55

• 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.
– Stiv
Oct 17 at 12:31