Assemble a formula using the numbers $2$, $0$, $1$, and $8$ in any order so that the result equals $199$. 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 $2$, $0$, $1$, or $8$. Operands may of course also be derived from calculations e.g. $10+(\sqrt{8*2})!$. You may also use brackets to clarify order of operations, and you may concatenate two or more of the four digits you start with (such as $2$ and $8$ to make the number $28$) if you wish. You may only use each of the starting digits once and you must use all four of them. I'm afraid that concatenation of numbers from calculations is not permitted, but answers with concatenations which get $199$ will get plus one from me.

Double, triple, etc. factorials (n-druple-factorials), such as $4!! = 4 \times 2$ are not allowed, but factorials of factorials are fine, such as $(4!)! = 24!$. I will upvote answers with double, triple and n-druple-factorials which get $199$, but will not mark them as correct. (This is particularly important because otherwise the method of @CarlSchildkraut in answer to this question could be used to get a solution with little effort).

Here are some examples to this problem:

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

This question is based on a previous question where the target was 109. Many thanks to @Hugh who edited and improved that question and others who made useful comments and suggestions...


1 Answer 1


How about:


  • $\begingroup$ well done - that was quick :-) $\endgroup$
    – tom
    Commented Sep 18, 2018 at 19:39
  • $\begingroup$ I spotted $199^2\approx8!$ and the rest just followed $\endgroup$
    – JMP
    Commented Sep 19, 2018 at 4:27
  • $\begingroup$ Yes, I see that now...., will have to think a bit more before the next one.... $\endgroup$
    – tom
    Commented Sep 19, 2018 at 8:11

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