Can you assemble a formula using the numbers $2$, $0$, $2$, $0$, and $2$ in any order so that the results equals $505$. You may use the operations $x + y$, $x - y$, $x \times y$, $x \div y$, $x!$, $\sqrt{x}$, $\sqrt[\leftroot{-2}\uproot{2}x]{y}$ and $x^y$, as long as all operands are either $2$, $0$, $2$, $0$, or $2$. Operands may of course also be derived from calculations e.g. $200*(2+2)$. You may also use brackets to clarify order of operations, and you may concatenate two or more of the five digits you start with (such as $2$ and $0$ to make the number $20$) if you wish. You may only use each of the starting digits once and you must use all five of them. I'm afraid that concatenation of numbers from calculations is not permitted, but answers with concatenations will get plus one from me.

Any finite number of functions can be used, though ingenious solutions with infinite numbers of functions will get plus one from me.

Note that
double, triple, etc. factorials (n-druple-factorials), such as $6!! = 6 \times 4 \times 2$ are not allowed, but factorials of factorials are fine, such as $((2+0!)!)! = 6! = 720$. I will upvote answers with double, triple and n-druple-factorials which get the required answers, but will not mark them as correct - particularly because a general method was developed by @Carl Schildkraut to solve these puzzles provided that you can make a single number greater than $2$ two times from the numbers you start with - here it could be $20+0$ and $20$ to get two $20$s, for example.

many thanks to the authors of the similar questions below for inspiring this question.


1 Answer 1


How about


How did I find it?

$504$ is equal to $24\times21$ and you can find $24$ with two factorials in a row with two $2$, but we cannot have 21 without using $2$ and $1$ at the same time, but we had 2 and 0. So the rest becomes suspicious since only 23 and 22 was missing in between which makes me close to 504 more and we could eliminate less than 20! with 2 and 0 and so on.

  • 3
    $\begingroup$ Good job :-) well done $\endgroup$
    – tom
    Sep 23, 2018 at 20:41

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