Make 1998 using the least possible digits 8

Thanks to a comment from Ben Barden, here is another way of achieving 11 8s

I have a solution with 12 8s

Updated, another with 9 8s

Stealing gloriously from the work of others, I have it down to 11:

A solution with nine $8$s:

i.e.

A very simple solution with ten $8$s (which I'm surprised nobody else has done):

Here's a solution with $9$ eights, without using the % operator:

My first try, with ten:

Only 4 operators

Found a solution with 8 eights, using concatenation and finally finding some use for the percent sign:

EDITED (much later..): Found another, without concatenation this time:

Here is a hilarious solution for 9

For research purposes I'll also include my kinda illegal solution for 7

If you allow concatenation of intermediate results (not just the original $8$s), here's a solution with $7$ eights:

OK, so I took a different approach. Seeing as I couldn't come up with anything interesting, I decided - f-it, let's make the computer try! And wrote a little program that tries all the possibilities. The code can be found here on PasteBin.

There are two things of note about the % operator:

• I treated it as an unary operator which divides by 100. So it can be stacked too: (8+8)%% = 0.0016
• Since you can potentially add as many % operators as you want to a single operand, I had to put in some kind of limit. Initially I set it to max 3 % operators in a row, but later changed to 1 to make it faster.

With that in mind the results are...

I couldn't find any expressions with 6 8s or less. But with 7 8s they started coming in. Here's one:

The total results for 7x8 with no more than 1 % in a row are below. They are all in Polish Notation because that was easier for me to produce. Converting them to "normal" notation is straightforward, but tedious, so I'll leave that to someone else. :)

Note: I've checked all formulas with 6x8 and up to 2 % signs in row and didn't find anything. I also checked all 5x8 with up to 3 % signs in row. No results. However this doesn't prove that it's impossible to do with 6 or less 8. This only means that my code cannot find such combinations because it's beyond what it is capable of. The double data type does have its limits, and Legorin showed that you can have a legit answer with 1875 % signs in row (which is awesome, by the way). The code could be further improved to both increase accuracy and speed, but I've already wasted enough time on it as it is. If you want to give it a go, be my guest! :)

I found another solution with 8.

the %%...%% is 1875 % symbols

therefore

Straightforward solution with 9 8s:

I can solve with nine 8s.

$(frac{8+8}{8}+8+8)$ $timesfrac{888}{8}$