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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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    Also, we may not split the ones or combine, so 1+1 and 1111 are not allowed.2012-11-30

5 Answers 5

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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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    @user1511417 - This is a bruteforced solution.2017-09-03
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$11+11+11\neq 6$

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

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    And $(11 \cdot 11 + 11)/(11+11)=6.$ Oops, too many 11's.2012-12-01
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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$