Can you assemble a formula using the least amount of one digit numbers (from $0$ to $9$) so that the results equals to 2018 with the rules below?

  • You may use the operations $x + y$, $x - y$, $x \times y$, $x \div y$, $x!$, $\sqrt[\leftroot{-2}\uproot{2}x]{y}$ and $x^y$,
  • Using a direct square root is not allowed since it is actually power of $0.5$.
  • You may use brackets to clarify order of operations.
  • You are allowed to use one digit number as much as you want, such as you may try to assembly a formula using four $2$s two $1$s etc.
  • You are not allowed to concatenate.
  • Double, triple, etc. factorials (n-druple-factorials), such as $4!! = 4 \times 2$ are not allowed either.
  • 1
    $\begingroup$ Easy. I can use the word form, "two thousand eighteen". $\endgroup$ – Alto Sep 22 '18 at 22:22

Update: I can do it using

$4$ digits


$(9\times 5)^2 - 7 = 2018$


I found a way to do it in

4 digits as well


$6^4 + 2 + 6!$


$8!/(4\times 5)+2$

  • $\begingroup$ I have edited the $\ast$ to $\times$ (generated by writing $\times$). If you disapprove, please let me know. $\endgroup$ – Feeds Sep 23 '18 at 0:53
  • $\begingroup$ If you want, you could write $\div$ to generate $\div$, but the slash / is mostly preferred :) $\endgroup$ – Feeds Sep 23 '18 at 1:03

already solved, but here is another solution

four digits... $${9! \over 6!} \times 4+2=2018$$

but this is similar to Excited @Raichu's second answer...


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