# How do I make 76 using only the numbers 2,0,1,9? [closed]

I must use each number 2,0,1,9 (only once) to come up with an answer of 76

• What operations are allowed? Can we concatenate numbers? – Glorfindel Dec 11 '18 at 15:15
• i believe any operations are allowed. i tried to spell out squared, but that was not allowed (2+0) squared * 19 = 76 they said the squared was an additional 2 – Chris Dec 11 '18 at 15:21
• @Chris Could you provide a source for this puzzle? Site policy dictates that you provide a source for a puzzle, in case it is not yours and evidently, you didn't come up with this puzzle on your own by your comments. – Sid Dec 11 '18 at 15:23
• It's not mine. it's a school assignment. we have to make numbers 1-100 using only 2019 – Chris Dec 11 '18 at 15:24
• Pretty sure most SE sites are against answering questions that are specifically for homework assignments. – Robert S. Dec 11 '18 at 15:26

$$\frac{9}{.\overline{1}} - \frac{0!}{.2} = 76$$

where

$$.\overline{1} = .111111\ldots$$

• I think this is the correct answer, well done. – TakingNotes Dec 11 '18 at 16:33

$$-\log_{\sqrt{9}!-2}(\log(\underbrace{\sqrt{\sqrt{...\sqrt{10}}}}_\text{158 square roots})$$ The "158" is not part of the equation. Normally, you'd write down all 158 square roots.

• One closing bracket is missing. Can you explain while this formula equals to 76 ? – Evargalo Dec 12 '18 at 16:41

(2+1)! + 0! = 7, concatenate with 9 flipped over = 76.

Or:

you used 19 in your guess, so I'm assuming concatenating the original numbers is allowed: (9+2)!!!!! + 10

• dang they won't allow flipping or concatenate. sorry, i just asked. thank you for trying :) – Chris Dec 11 '18 at 15:23
• Your second guess comes out to be 65. – JR_M Dec 11 '18 at 15:54
• @JR_M I’ve deleted one exclamation point, it should work now. – Excited Raichu Dec 11 '18 at 15:55
• sorry, 65. I think you have one too many factorials – JR_M Dec 11 '18 at 15:57
• – Excited Raichu Dec 11 '18 at 15:59

I was doing a bit of research into factorials, and found both hyperfactorials (denoted by an $$H$$) and alternating factorials (denoted by an $$AF$$). Hopefully this answer fulfills your need.

$$AF(\sqrt{9} + 1) * H(2 + 0)$$

First we can take the hyperfactorial of $$H(2 + 0)$$.

$$AF(\sqrt{9} + 1) * 4$$

We'll then solve for the other pair of brackets.

$$AF(4) * 4$$

Now we'll take the alternating factorial.

$$19 * 4$$

And then some basic multiplication to get:

$$76$$

• Isn't the square root implicitly using an extra digit 2? Because sqrt that is (x)^1/2 – rhsquared Dec 11 '18 at 16:23
• By the Pickover definition, the superfactorial of 3 is about 10^10^10^10^36000. By the Sloane/Plouffe definition, the superfactorial of 3 is 12, or if you're counting 0, 2. Neither of them are 72. – Excited Raichu Dec 11 '18 at 16:30
• Yes, thank you for pointing out my error. I had taken the Sloane/Pouffe definition of superfactorial(4) and divided it by 4, not 24. Thankyou for pointing that out. – TakingNotes Dec 11 '18 at 16:32

(9^2)-(2^2)-(1^2)-(0^2)=76
If we take the square of all the numbers and apply subtraction then we get 76.

• Sorry to burst your bubble, but you cant repeat any numbers. – TakingNotes Dec 11 '18 at 15:47
• According to OP's first comment, if you do a number's square you're already using up the 2. – S. M. Dec 11 '18 at 15:48
• Also, don't forget to hide your answers using ">!" at the beginning of a line – eye_am_groot Dec 11 '18 at 15:49

$$(((\sqrt{9})!)!!-10)*2 = 76$$

Here is the answer! Finally!!!!! And thanks to all who helped in the spirit of solving a puzzle.

• My second answer is just as valid as this. – Excited Raichu Dec 11 '18 at 16:24
• @Chris, you said concatenated numbers weren't valid. – S. M. Dec 11 '18 at 16:25
• My answer is surprisingly even more valid than this. – TakingNotes Dec 11 '18 at 16:27