Fewest digit 8s and +-*/() to get 1000?

Question

Here it is with eight 8s: $$888+88+8+8+8 = 1000$$ Here it is with seven 8s: $$8\times8\times(8+8)-8-8-8 = 1000$$

Is it possible with less?

Answer 1

There are only $3$ essentially different solutions with $7$ digits "$8$":

$$ (8+8)\cdot 8 \cdot 8 - 8 - 8 - 8 = 1000; $$

$$ 8\cdot (8+8\cdot(8+8))-88 = 1000; $$

$$ (8+8)\cdot \left(8\cdot8 -\frac{8}{8}\right)-8 = 1000. $$

To find them, one can build:

1 - list of possible numbers, which can be obtained from $1$ digit "$8$": $\{8\}$;

2 - list of possible numbers, which can be obtained from $2$ digits "$8$": $\{0,1,16,64,88\}$;
etc.
(Each list is derived from all previous ones).

And $6$th list will be without number $1000$, so there are no solutions with $6$ (or less) digits "$8$".
If only $+,-,*,/$ and $()$ are allowed.