# Use 2, 0, 1 and 8 to make 71

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.

• how is 12!!!!!!=12×6? – Rotsor Sep 9 '18 at 13:32
• @Rotsor Sextuple factorial (not allowed in this puzzle). – EKons Sep 9 '18 at 18:10
• @ΈρικΚωνσταντόπουλος, it's just a question how does it work, mathematically. – rus9384 Sep 10 '18 at 7:39
• @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. ;-) – EKons Sep 10 '18 at 9:11
• Is $(N!)!$ allowed? – rus9384 Sep 10 '18 at 9:40

$$\sqrt{(8-1)!+2-0!}$$ I found this by chance when noticing that $71^2 = 5041$ was extremely close to $7! = 5040$.
$$.1\times(8-2)!-0! = .1\times6!-1 = 72-1 = 71$$