# Use the digits 1, 9, 9, 8 to get 81 [closed]

Using the digits $$1, 9, 9$$ and $$8$$ in that order, and the operations $$+$$, $$-$$ and/or $$\times$$ in any order between pairs of digits, create an expression that equals $$81$$.  Normal precedence rules apply.

As an example, here is one solution: $$1 \times 9 + 9 \times 8 = 81$$. Find three or four more solutions that fit the criteria.

• Paris of digits? – zagad Sep 21 '18 at 10:47
• @zagad: Probably a typo of "pairs of digits". – Rudy Velthuis Sep 21 '18 at 10:50
• @RudyVelthuis Oh, right! I just couldn't figure out what the auto-corrector corrected here :) – zagad Sep 21 '18 at 10:51
• @KeykoYume: But ++ is a different operator than +. And - in -1 is the unary negation operator, not sure if that is allowed either. – Rudy Velthuis Sep 21 '18 at 11:11
• It is unclear what is allowed. Are parentheses allowed? What about and or? Are only $+ - \times$ allowed? You ask for "three of four more solutions." Do they exist? – Weather Vane Sep 21 '18 at 12:19

If truly only + - X are accepted, no parenthesis or concatenations, and the numbers must stay in this order, all I can see for now are those.

1 x 9 + 9 x 8 = 81
-1 x -9 + 9 x 8 = 81
1 x 9 + -9 x -8 = 81
-1 x -9 + -9 x -8 = 81

• Well done I completely forgot about using negatives. Not sure if its what they wanted but it works! – gabbo1092 Sep 21 '18 at 13:04
• That's most simple and correct answer, well done – Jakhu AE Sep 22 '18 at 11:35
• Ah, I didn't think of using negatives, because the question says "in any order between pairs of digits"... I guess if the OP says it's correct, then it is! – Hugh Sep 22 '18 at 17:01

Not sure if this is considered cheating or not but would these solutions count?

((-1-9)*9-8)--

or:

((-1+9)*9+8)++

or:

((19-9)*8)++

• Probably not: ++ and -- are not the same operators as + and -. IOW, it is probably cheating and invalid. – Rudy Velthuis Sep 21 '18 at 11:12
• @RudyVelthuis yeah, I think you're right – KeykoYume Sep 21 '18 at 11:13

Another answer to this question is:

9*8*1+9

I'm also pretty sure that that is the only other answer. Here's why:

To get 81, we need to get any of the following: 80, 81, or 82 with 9,9,8; 9, 72, or 90 with 1,8,9; or 73 or 89 with 9,9,1. Let's start with 9,9,1. The closest we can come to 73 is 9*(9-1) = 72. Likewise, we also miss 89 by one, with 9*(9+1) = 90. Any other combination either gives a number in the 80s or number that is too small. Continuing with 1,8,9, it is possible to get 72 in one of the two ways (that we have already said.) Getting 9 is impossible; your numbers are either too high or too low (9+1-8) = 2, 9+8-1 = 16). The highest possible number to get by using 1,8,9 with your rules is 9*(8+1) = 81, so 90 is out of the question. Finally, 9, 9, 8. At first glance, it seems possible to get 81 (9*8+9), but this solution is the same as the solutions from 72 with 1,8,9. Also, it's not possible to get 80 or 82, because the only way to get close is 9*9 or 9*8. 9*9 is 81, which + or -8 doesn't help; and 9*8 + 9 is 81, which we've covered. So, the only two (eight with negative numbers) possible solutions with your rules are 9*8*1+9 and 9*8+9*1.

• Well done! The question does technically says that you have to keep the order of the digits as 1-9-9-8 – Hugh Sep 21 '18 at 22:41

I am assuming by 'and or' you mean bitwise and or

Following is a possible solution:

1 or 9 x 9 or 8=81, bitwise has higher precedence than x

• lol, but it doesn't have the lateral-thinking tag so I doubt that this is the correct answer. – Hugh Sep 21 '18 at 22:42