I must use each number 2,0,1,9 (only once) to come up with an answer of 76
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1$\begingroup$ What operations are allowed? Can we concatenate numbers? $\endgroup$– GlorfindelCommented Dec 11, 2018 at 15:15
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$\begingroup$ 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 $\endgroup$– ChrisCommented Dec 11, 2018 at 15:21
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1$\begingroup$ @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. $\endgroup$– SidCommented Dec 11, 2018 at 15:23
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$\begingroup$ It's not mine. it's a school assignment. we have to make numbers 1-100 using only 2019 $\endgroup$– ChrisCommented Dec 11, 2018 at 15:24
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4$\begingroup$ Pretty sure most SE sites are against answering questions that are specifically for homework assignments. $\endgroup$– Robert S.Commented Dec 11, 2018 at 15:26
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6 Answers
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How about this
$\frac{9}{.\overline{1}} - \frac{0!}{.2} = 76$
where
$.\overline{1} = .111111\ldots$
answered Dec 11, 2018 at 16:22
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$\begingroup$ I think this is the correct answer, well done. $\endgroup$ Commented Dec 11, 2018 at 16:33
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$$-\log_{\sqrt{9}!-2}(\log(\underbrace{\sqrt{\sqrt{...\sqrt{10}}}}_\text{158 square roots})$$ The "158" is not part of the equation. Normally, you'd write down all 158 square roots.
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$\begingroup$ One closing bracket is missing. Can you explain while this formula equals to 76 ? $\endgroup$– EvargaloCommented Dec 12, 2018 at 16:41
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(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
answered Dec 11, 2018 at 15:22
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1$\begingroup$ dang they won't allow flipping or concatenate. sorry, i just asked. thank you for trying :) $\endgroup$– ChrisCommented Dec 11, 2018 at 15:23
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$\begingroup$ Your second guess comes out to be 65. $\endgroup$– JR_MCommented Dec 11, 2018 at 15:54
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$\begingroup$ @JR_M I’ve deleted one exclamation point, it should work now. $\endgroup$ Commented Dec 11, 2018 at 15:55
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$\begingroup$ sorry, 65. I think you have one too many factorials $\endgroup$– JR_MCommented Dec 11, 2018 at 15:57
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1$\begingroup$ @S. M. en.m.wikipedia.org/wiki/Factorial#Multifactorials $\endgroup$ Commented Dec 11, 2018 at 15:59
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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:
$76$
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$\begingroup$ Isn't the square root implicitly using an extra digit 2? Because sqrt that is (x)^1/2 $\endgroup$ Commented Dec 11, 2018 at 16:23
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$\begingroup$ 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. $\endgroup$ Commented Dec 11, 2018 at 16:30
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$\begingroup$ 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. $\endgroup$ Commented Dec 11, 2018 at 16:32
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(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.
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3$\begingroup$ Sorry to burst your bubble, but you cant repeat any numbers. $\endgroup$ Commented Dec 11, 2018 at 15:47
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1$\begingroup$ According to OP's first comment, if you do a number's square you're already using up the 2. $\endgroup$– S. M.Commented Dec 11, 2018 at 15:48
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$\begingroup$ Also, don't forget to hide your answers using ">!" at the beginning of a line $\endgroup$ Commented Dec 11, 2018 at 15:49
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$(((\sqrt{9})!)!!-10)*2 = 76$
Here is the answer! Finally!!!!! And thanks to all who helped in the spirit of solving a puzzle.
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2$\begingroup$ My second answer is just as valid as this. $\endgroup$ Commented Dec 11, 2018 at 16:24
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2$\begingroup$ @Chris, you said concatenated numbers weren't valid. $\endgroup$– S. M.Commented Dec 11, 2018 at 16:25
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2$\begingroup$ My answer is surprisingly even more valid than this. $\endgroup$ Commented Dec 11, 2018 at 16:27