## Friday, August 9, 2013

### Solving a Bar Bet Using at Most Three Different Operations

A bar bet as presented in the YouTube video The HARDEST Puzzle Yet! involves starting with three of the same number from 0 through 9, then adding mathematical operations that result in an evaluation of 6. For example, if we start with three of the number 6, one solution could be $6+6-6=6 .$ I'll demonstrate a method for solving this bar bet puzzle starting with three of any number, say $$N$$, which involves using at most three different mathematical operations (although some of these may be used many, many times).

If $$0 \leq N \leq 2$$ we have the solutions
\begin{eqnarray}
(0! + 0! + 0!)! &= 6 \\
(1! + 1! + 1!)! &= 6 \\
2+2+2 &= 6.
\end{eqnarray} If $$N\geq 3$$, concatenate the three numbers. Repeatedly applying the square-root operation we'll eventually end up with a result $$x$$ with $$3 \leq x < 9$$.  If we now take the greatest integer $$\lfloor x \rfloor$$ we have an integer $$n$$ with $$3 \leq n \leq 8$$. If we can exhibit solutions for each of these cases that use only square-roots, greatest integers and one other operation, we'll be done. Using factorial for the third operation, some possibilities are
\begin{eqnarray}
\href{http://www.wolframalpha.com/input/?i=3%21}{3!} &= 6\\
\href{http://www.wolframalpha.com/input/?i=%5Cleft%5Clfloor+%5Csqrt%7B%5Csqrt%7B%5Cleft%5Clfloor+%5Csqrt%7B%5Csqrt%7B%5Csqrt%7B%5Csqrt%7B%5Csqrt%7B%284%21%29%21%7D%7D%7D%7D%7D+%5Cright%5Crfloor%21%7D%7D+%5Cright%5Crfloor+%21}{\left\lfloor \sqrt{\sqrt{\left\lfloor \sqrt{\sqrt{\sqrt{\sqrt{\sqrt{(4!)!}}}}} \right\rfloor!}} \right\rfloor !} &= 6\\
\href{http://www.wolframalpha.com/input/?i=%5Cleft%5Clfloor+%5Csqrt%7B%5Csqrt%7B5%21%7D%7D+%5Cright%5Crfloor+%21}{\left\lfloor \sqrt{\sqrt{5!}} \right\rfloor !} &= 6\\
\href{http://www.wolframalpha.com/input/?i=6}{6} &= 6 \\
\href{http://www.wolframalpha.com/input/?i=%5Cleft%5Clfloor+%5Csqrt%7B%5Csqrt%7B%5Csqrt%7B%5Cleft%5Clfloor+%5Csqrt%7B%5Csqrt%7B7%21%7D%7D%5Cright%5Crfloor+%21%7D%7D%7D+%5Cright%5Crfloor+%21}{\left\lfloor \sqrt{\sqrt{\sqrt{\left\lfloor \sqrt{\sqrt{7!}}\right\rfloor !}}} \right\rfloor !} &= 6\\
\href{http://www.wolframalpha.com/input/?i=%5Cleft%5Clfloor+%5Csqrt%7B%5Csqrt%7B%5Csqrt%7B8%21%7D%7D%7D+%5Cright%5Crfloor+%21}{\left\lfloor \sqrt{\sqrt{\sqrt{8!}}} \right\rfloor !} &= 6.
\end{eqnarray}
I've added Wolfram Alpha links so you can verify that these do indeed evaluate to 6.

As an illustrative example, when $$N=1337$$ we have the solution $\href{http://www.wolframalpha.com/input/?i=%5Cleft%5Clfloor+%5Csqrt%7B%5Csqrt%7B%5Cleft%5Clfloor+%5Csqrt%7B%5Csqrt%7B%5Csqrt%7B%5Csqrt%7B%5Csqrt%7B%28++%5Cleft%5Clfloor+%5Csqrt%7B%5Csqrt%7B%5Csqrt%7B%5Csqrt%7B133713371337%7D%7D%7D%7D+%5Cright%5Crfloor+%21%29%21%7D%7D%7D%7D%7D+%5Cright%5Crfloor%21%7D%7D+%5Cright%5Crfloor+%21}{\left\lfloor \sqrt{\sqrt{\left\lfloor \sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\left( \left\lfloor \sqrt{\sqrt{\sqrt{\sqrt{133713371337}}}} \right\rfloor !\right)!}}}}} \right\rfloor!}} \right\rfloor !} = 6.$