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.

  • 2
    $\begingroup$ how is 12!!!!!!=12×6? $\endgroup$ – Rotsor Sep 9 '18 at 13:32
  • $\begingroup$ @Rotsor Sextuple factorial (not allowed in this puzzle). $\endgroup$ – EKons Sep 9 '18 at 18:10
  • $\begingroup$ @ΈρικΚωνσταντόπουλος, it's just a question how does it work, mathematically. $\endgroup$ – rus9384 Sep 10 '18 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 '18 at 9:11
  • $\begingroup$ Is $(N!)!$ allowed? $\endgroup$ – rus9384 Sep 10 '18 at 9:40

How about:

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

  • $\begingroup$ Please hide your hint. $\endgroup$ – Hack Saw Sep 9 '18 at 8:58
  • 1
    $\begingroup$ @HackSaw Do you mean like this? $\endgroup$ – Toby Mak Sep 9 '18 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 '18 at 18:50
  • $\begingroup$ @LegionMammal978 Wow, thanks for confirming! I must have been extremely lucky to find this. $\endgroup$ – Toby Mak Sep 9 '18 at 22:24
  • 1
    $\begingroup$ 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 '18 at 0:24

$$.1\times(8-2)!-0! = .1\times6!-1 = 72-1 = 71$$

  • $\begingroup$ 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 '18 at 8:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.