# Find 2018 with the least amount of numbers

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.
• Easy. I can use the word form, "two thousand eighteen". – 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

Using

$$6^4 + 2 + 6!$$

Or:

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

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

These use more digits than the pre-existing answers do but I think are worth mentioning anyway:

$$3^7-(6+7)^2$$ $$2^{5+6}-5\times6$$