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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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    Can I use an extra 5 please? It would make it much easier2012-11-30
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    Presumably you don't like $1+1+1+1+1+1=6$2012-11-30
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    It is easy to do in octal numeral system. :)2012-11-30
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    What is (,) operation?2012-11-30
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    If the square root operator is in fact $\lfloor\sqrt{\bullet}\rfloor$ then this is trivial, otherwise it looks impossible.2012-11-30
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    Yeah, integer arithmetic can solve this. I still think that the to me mysterious , operator might help. Eg, if x=ab (digits of x) and ,(x) = (a,b), then (1,1) + (1,1) + (1,1) = (3,3) which is getting close.2012-11-30
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    Sorry, I did not intend (,) as an operation, but as symbols we may put into the equation. That is, we may turn $5~~5~~5$ into $5-(5+5)$2012-11-30
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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

11

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

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    +1 I suppose the triple factorial is not among the allowed operators, but nevertheless, this solution rox!2012-11-30
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    @dtldarek - Set of available symbols is limited, but there are no restrictions imposed on available operators established by OP. :)2012-11-30
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    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 ;-)2012-11-30
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    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 ;-)2014-09-03
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    How? How did you come up with this solution, Egor?? Which buttons do you have to push in ones brain to get there? I do not get it.2017-09-03
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    @user1511417 - This is a bruteforced solution.2017-09-03
6

$11+11+11\neq 6$

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:

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