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Use all and only the digits $2,0,1,8$ once each to make the number $71$.

Allowed operations; anything not on this list is banned:

  • $+,-,\times,\div, ()$ (parentheses and/or choose function)

  • Concatenation; only applied to the original digits e.g $(8-1)||(2-0!)$ is not allowed

  • $!$ single factorial (none of that double factorial + weird stuff otherwise you could do something like $12!!!!!!=12\times6$ and that's a bit cheat)

  • Exponentiation, although the exponent must be 'made' as well

  • Sqrt (free of cost); nth roots however require you to be able to make the number 'n'

  • Decimal point: like concatenation, this can only be applied to the original digits. Sorry to those who attempted this before -- unlike in some questions, I'm requiring that any decimal point needs an integer part before it (wikipedia: ..used to separate the integer part from the fractional part of a number)

Sorry I know PSE is being plagued with these but I couldn't resist.

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    $\begingroup$ how is 12!!!!!!=12×6? $\endgroup$
    – Rotsor
    Sep 9, 2018 at 13:32
  • $\begingroup$ @Rotsor Sextuple factorial (not allowed in this puzzle). $\endgroup$
    – EKons
    Sep 9, 2018 at 18:10
  • $\begingroup$ @ΈρικΚωνσταντόπουλος, it's just a question how does it work, mathematically. $\endgroup$
    – rus9384
    Sep 10, 2018 at 7:39
  • $\begingroup$ @rus9384 It's a matter of (totally wrong, in my opinion) notation; don't try to split the $!$s to find some meaning which makes more sense. ;-) $\endgroup$
    – EKons
    Sep 10, 2018 at 9:11
  • $\begingroup$ Is $(N!)!$ allowed? $\endgroup$
    – rus9384
    Sep 10, 2018 at 9:40

2 Answers 2

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How about:

$$\sqrt{(8-1)!+2-0!} $$ I found this by chance when noticing that $71^2 = 5041$ was extremely close to $7! = 5040$.

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  • $\begingroup$ Please hide your hint. $\endgroup$
    – Hack Saw
    Sep 9, 2018 at 8:58
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    $\begingroup$ @HackSaw Do you mean like this? $\endgroup$
    – Toby Mak
    Sep 9, 2018 at 9:28
  • $\begingroup$ I tried brute-forcing this using my computer, and this solution (including hundreds of variations) was the only one it could find using the rules. $\endgroup$
    – LegionMammal978
    Sep 9, 2018 at 18:50
  • $\begingroup$ @LegionMammal978 Wow, thanks for confirming! I must have been extremely lucky to find this. $\endgroup$
    – Toby Mak
    Sep 9, 2018 at 22:24
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    $\begingroup$ Yeah just some background to this -- (spoiler alert if you follow the link) which is kinda cool -- en.wikipedia.org/wiki/Brocard%27s_problem $\endgroup$
    – Wen1now
    Sep 10, 2018 at 0:24
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$$.1\times(8-2)!-0! = .1\times6!-1 = 72-1 = 71$$

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  • $\begingroup$ Hi there, welcome to Puzzling :) I'm real sorry -- I forgot to specify that the decimal point needs an integral part before it. +1 anyway $\endgroup$
    – Wen1now
    Sep 9, 2018 at 8:08

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