# 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

## closed as too broad by Christoph, AHKieran, w l, JonMark Perry, rhsquaredDec 11 '18 at 16:26

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• 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. – a guy 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

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. – a guy Dec 11 '18 at 16:32

(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

(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. – a guy 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 – Greg 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. – a guy Dec 11 '18 at 16:27