# Use three 11's and various math symbols to make an equation equal to 6

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 easier – Adam Rubinson Nov 30 '12 at 15:44
Presumably you don't like $1+1+1+1+1+1=6$ – Ross Millikan Nov 30 '12 at 16:12
It is easy to do in octal numeral system. :) – Egor Skriptunoff Nov 30 '12 at 16:29
What is (,) operation? – gnometorule Nov 30 '12 at 16:48
If the square root operator is in fact $\lfloor\sqrt{\bullet}\rfloor$ then this is trivial, otherwise it looks impossible. – dtldarek Nov 30 '12 at 17:26

## 4 Answers

$\large 6=\left( \sqrt{\sqrt{\frac{11+11!!!}{11}}}\right)\LARGE!$
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! – dtldarek Nov 30 '12 at 18:08
@dtldarek - Set of available symbols is limited, but there are no restrictions imposed on available operators established by OP. :) – Egor Skriptunoff Nov 30 '12 at 18:27
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 ;-) – dtldarek Nov 30 '12 at 18:43
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 ;-) – chiastic-security Sep 3 '14 at 12:11

$11+11+11\neq 6$

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. – coffeemath Dec 1 '12 at 15:12

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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If you can use log to mean log base 2 (or lb which is an ISO symbol for log base 2 although rarely used), then you can do it this way:

-log((log(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt((sqrt 11*sqrt 11)))))))))/(log(11))) = 6


This link to Google (using log2 so that google understands it) shows that it works.

This method is based on a trick that the great Paul Dirac came up with whilst a student at Cambridge. You can get to any answer by changing the number of 'sqrt's in the outside bit (there are six above so the answer is 6) and you can substitute any number in place of 11 as this gets cancelled out anyway.

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