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How can one make $66$ with only $1,1,1,1,1$? You cannot combine these two numbers to make a new number, such as this: $66=11 \times (1+1+1)!$.

This was inspired a game of dice that I used to play, where you were given $5$ numbers between $1$ and $6$ to make a two digit number between $11$ and $66$. My mother had once told me that no matter which numbers were given, you could use operations such as $\times$, $\div$,$+$,$-$, $!$, $x^y$, $\sqrt [ x ]{ y }$, $\sqrt { y }$, and %, ‰ to make a equation. So naturally I tried what seemed to most difficult to make: making $66$ with $1,1,1,1,1$.

However, I was unable to find a solution. I initially thought that this would be impossible, but considering that it is possible to make $97$ from $0,0,0,0$ I suspect that there might be a way.

Is there a possibly easy solution to this problem?

(Note: Making $97$ with $0,0,0,0$, can be seen here, on the third comment to David Bevan`s Answer, by David Bevan.)

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Chad Shin
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  • We can at least get an upper bound of six $1$'s: $\left(\frac{1}{\sqrt{1 %}} + 1\right) (1 + 1 + 1)!$. – Travis Willse Feb 03 '16 at 01:58
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    If one permits the floor function $\lfloor,\cdot,\rfloor$, we can write $$\left\lfloor \frac{1 + 1}{\left(1 + 1 + 1\right)%}\right\rfloor .$$ – Travis Willse Feb 03 '16 at 02:21

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