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Use 2 0 1 and 8 to make the number 67

RULES

  1. You must use all 4 digits. Only the digits 2, 0, 1, and 8 can be used. You can make multi-digit numbers out of the numbers. Examples: 20, 82, 2.8

  2. The square function may NOT be used. Nor may the cube, raise to a fourth power, or any other function that raises a number to a specific power. You may use the ^ operation if you use a digit, for example, [(8 + 1)^2 - 0!] is acceptable (if you're trying to get 80), because 2, 0, 1, and 8 is used. However, [20 ^ 2 / 8 + 1] can't be used to get 51 because it uses an extra 2.

  3. Sorry, but the integer function may NOT be used. Nor may the round, floor, ceiling, or truncate functions.

  4. +, -, *, /, (), !, sqrt, ^, and !! may be used for functions.

Please no brute-force methods. Good luck.

I see that there are many answers. I like #3 because it doesn't use more than 2 factorials, but the most upvoted one gets the credit.

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    $\begingroup$ $8^2+0!+1=66$ dammit! $\endgroup$ Commented Sep 9, 2018 at 0:12
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    $\begingroup$ @user477343 also $8^2 + 0 + 1 = 65$, $80-12=68$, $8^2 + (1 \times 0) = 64$, $8^2 - 0 - 1 = 63$, and $8^2 - 0! - 1 = 62$ haha!! $\endgroup$ Commented Sep 9, 2018 at 0:59
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    $\begingroup$ Fun fact! $\frac{8!}{20}+1=2017$ $\endgroup$ Commented Sep 9, 2018 at 2:14
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    $\begingroup$ Out of curiosity, why 67? $\endgroup$ Commented Sep 9, 2018 at 3:47
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    $\begingroup$ It makes no sense to permit sqrt but disallow square. $\endgroup$ Commented Sep 9, 2018 at 8:04

6 Answers 6

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How about:

$(8!!!!+1)\times2+0!\\=(8\cdot4+1)\times2+1\\=(32+1)\times2+1\\=33\times2+1\\=66+1\\=67$

and similarly, with the digits in order:

$20!!!!!!!!!!!!!!!!! - 1 + 8$, where $20!^{(17)}=20\cdot3=60$.

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    $\begingroup$ I think $8!! = 8 \times 6 \times 4 \times 2$ $\endgroup$ Commented Sep 8, 2018 at 22:28
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    $\begingroup$ @Marius I think this is correct based on the definition of the quadruple factorial; it’s up to OP to decide whether the quadruple factorial is allowed as a valid operation for this puzzle. Great answer though, JonMark Perry!! Love it! :D $\endgroup$ Commented Sep 8, 2018 at 22:33
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    $\begingroup$ 8!!!! is a multi-factorial $\endgroup$ Commented Sep 8, 2018 at 22:33
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$$8^2 + \sqrt{\frac{0!}{.\bar{1}}} = 64 + \sqrt{\frac{1}{\frac{1}{9}}} = 64 + 3 = 67$$ Note that $.\bar{1}$ is 0.111111... recurring.

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    $\begingroup$ Hello! Welcome to the Puzzling Stack Exchange (Puzzling.SE) and great answer, too! I see that you are familiar with the Stack Exchange community since you have been given the $+100$ reputation bonus (and correct me if I'm wrong, but have I seen you on Math.SE?); though I still suggest you visit the Help Center for more information (particularly here then here in the Asking section as you have not asked a question yet). Happy Puzzling! :D $\endgroup$ Commented Sep 9, 2018 at 5:21
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    $\begingroup$ Nice answer! Welcome to Puzzling. $\endgroup$ Commented Sep 9, 2018 at 8:06
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$8!!! - 12 - 0! = 8 \times 5 \times 2 - 12 - 1 = 80-13 = 67$

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    $\begingroup$ that is brilliant, but is your function after the 8 allowed??? $\endgroup$ Commented Sep 8, 2018 at 22:41
  • $\begingroup$ don't know. I guess we'll find out $\endgroup$ Commented Sep 8, 2018 at 22:42
  • $\begingroup$ plus one and we will see :-) didn't know that was possible... $\endgroup$ Commented Sep 8, 2018 at 22:42
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I base my answer on

75 (20 minus 18 in base 75 is 67 when converted back to decimal.)

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(2+1)! [connects (multi-digit) with] 8-0!
3! [connects (multi-digit) with] 8-1
3*2 [connects (multi-digit) with] 7
6 [connects (multi-digit) with] 7
67

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How about 2, 0 + 1, 8. If 2, 0 is in base 16 and 1, 8 is in base 27, it adds to 67 in base 10.

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    $\begingroup$ Why the downvotes? $\endgroup$ Commented Sep 9, 2018 at 21:25

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