Let's make a huge number with only tiny numbers

Question

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.

Answer 1

My answer is

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

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

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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 $

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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 $

Answer 2

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):

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

Answer 3

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, $55$?

Answer 4

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

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

⸂⸂⸜(രᴗര๑)⸝⸃⸃ woohoo

Answer 5

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

with 55 total square root signs.

Answer 6

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:

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

Answer 7

Using one tetration:

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

Answer 8

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$

Answer 9

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 $

Answer 10

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

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

Answer 11

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.

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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:

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

Answer 12

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

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

Answer 13

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

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

$= 8 - 1 + 16 + 32$

$= 7 + 16 + 32$

$= 7 + 48$

= 55