# Make numbers 93 using the digits 2, 0, 1, 8

I have a 1 to 100 building challenge with 2, 0, 1, 8, all made but 93 is not made. I need your help!

This is similar to the "Four fours" puzzle, but using the digits 2, 0, 1 and 8.

Rules:

• Use all four digits exactly once

• Allowed operations: +, -, x, ÷, ! (factorial), exponentiation, square root Parentheses and grouping (e.g. "21") are also allowed

• Squaring uses the digit 2 so expressions using multiple twos, like 2222 or 12+8212+82, are not allowed

• Multi-digit numbers and decimal points can be used such as 20, 102, .02 but you CANNOT make 30 by combining (2+1)0

• Recurring decimals can be used using the '

For example :

.2' = 0.222222.....

.12' = 0.1222222.....

.1'2' = 0.12121212.....

• Does square root also consume the 2? Jan 4, 2018 at 15:40

I think this is valid by your rules :

$.1^{-2} - 8 + 0!$
$= 100 - 8 + 1$
$= 93$

• Very nice! How long did it take you to come up with it? Jan 4, 2018 at 18:33
• I can't really tell since I was working in parallel^^. I just figured 10^2 - 8 was 92 and asked myself if I could get rid of the 0 for later use Jan 4, 2018 at 18:55
• Oh, my god! This is crazy! I've been thinking about it for more than a week, but it didn't work out. But I can finally finish the 2018 Challenge! Thank you very much! Jan 5, 2018 at 6:21

I am so close with: $\sqrt{\sqrt{\sqrt{12!}}}+80 \approx 92.16$

• Not using factorials or roots, $10^2-8=92$ Jan 4, 2018 at 15:54
• Oh, if only you could use the ceiling function! Jan 5, 2018 at 3:23