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The puzzle is to use the following symbols $$+,\;-,\;*,\;/,\;(\;,\;),\;!, \;\sqrt(\cdot)$$ in order to make a valid equation out of $$11~~~~~~11~~~~~~~11 = 6.$$

(There are three elevens with space in between for symbols).

This is part of a general series of questions about using any three integers in place of the elevens, but this case has me stumped.

So the question is to determine if it is possible to form a valid equation or how to prove it is not possible in an elegant way that avoids checking all possible cases.

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    $\begingroup$ Can I use an extra 5 please? It would make it much easier $\endgroup$ Nov 30, 2012 at 15:44
  • $\begingroup$ Presumably you don't like $1+1+1+1+1+1=6$ $\endgroup$ Nov 30, 2012 at 16:12
  • $\begingroup$ It is easy to do in octal numeral system. :) $\endgroup$ Nov 30, 2012 at 16:29
  • $\begingroup$ What is (,) operation? $\endgroup$ Nov 30, 2012 at 16:48
  • $\begingroup$ If the square root operator is in fact $\lfloor\sqrt{\bullet}\rfloor$ then this is trivial, otherwise it looks impossible. $\endgroup$
    – dtldarek
    Nov 30, 2012 at 17:26

6 Answers 6

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$\large 6=\left( \sqrt{\sqrt{\frac{11+11!!!}{11}}}\right)\LARGE!$
where n!!! = n(n-3)(n-6)... is triple factorial

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  • $\begingroup$ +1 I suppose the triple factorial is not among the allowed operators, but nevertheless, this solution rox! $\endgroup$
    – dtldarek
    Nov 30, 2012 at 18:08
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    $\begingroup$ @dtldarek - Set of available symbols is limited, but there are no restrictions imposed on available operators established by OP. :) $\endgroup$ Nov 30, 2012 at 18:27
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    $\begingroup$ Well, then for any $n$ you can define $(n-2)$-factorial, and have $((n*(n-(n-2))+n)/n)! = 6$, in particular $6 = \frac{11!!!!!!!!!+11}{11}!$. I still doubt this is a proper solution, but I like what you did ;-) $\endgroup$
    – dtldarek
    Nov 30, 2012 at 18:43
  • $\begingroup$ If you're going to define new symbols, it might just be easier to define $n!! = 6$ for all $n$ and have done with it ;-) $\endgroup$ Sep 3, 2014 at 12:11
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    $\begingroup$ @user1511417 - This is a bruteforced solution. $\endgroup$ Sep 3, 2017 at 11:40
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$11+11+11\neq 6$

EDIT: Hmm, I suppose that isn't strictly an "equation".

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  • $\begingroup$ And $(11 \cdot 11 + 11)/(11+11)=6.$ Oops, too many 11's. $\endgroup$
    – coffeemath
    Dec 1, 2012 at 15:12
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How about

$\sqrt{(11/11)/.\overline{11}}~!$

I know you didn't mention the bar symbol, but imo, it is a pretty standard mathematical operation.

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Binary 11 x Binary 11 - Binary 11 = 6 In computing, binary 00 = digital 0, 01 = 1, 10 = 2, 11 = 3

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Pretty amazing how the answers given so far miss the simplest possibility:

$$ \color{red} {\big(} \color{red} ( 1 \color{red} * 1 \color{red} ) \color{red} + \color{red} ( 1 \color{red} * 1 \color{red} ) \color{red} + \color{red} ( 1 \color{red} * 1 \color{red} ) \color{red} {\big)} \color{red} ! = 6$$

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If we're allowed multifactorials as in @Egor's answer, then we can make any $n > 3$ via $$ \left(\tfrac{\sqrt{n\times n}\underbrace{!!!\cdots !}_{n-3 \text{ times}}} { n} \right) !$$

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