15
$\begingroup$

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.

Improve this question
$\endgroup$
12
  • 1
    $\begingroup$ $8^2+0!+1=66$ dammit! $\endgroup$
    – Mr Pie
    Commented Sep 9, 2018 at 0:12
  • 1
    $\begingroup$ @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!! $\endgroup$
    – El-Guest
    Commented Sep 9, 2018 at 0:59
  • 4
    $\begingroup$ Fun fact! $\frac{8!}{20}+1=2017$ $\endgroup$
    – Bennett Bernardoni
    Commented Sep 9, 2018 at 2:14
  • 3
    $\begingroup$ Out of curiosity, why 67? $\endgroup$
    – user1717828
    Commented Sep 9, 2018 at 3:47
  • 1
    $\begingroup$ It makes no sense to permit sqrt but disallow square. $\endgroup$
    – Vicky
    Commented Sep 9, 2018 at 8:04

6 Answers 6

Reset to default
15
$\begingroup$

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 $20!^{(17)}=20\cdot3=60$.

Improve this answer
$\endgroup$
3
  • 4
    $\begingroup$ I think $8!! = 8 \times 6 \times 4 \times 2$ $\endgroup$
    – El-Guest
    Commented Sep 8, 2018 at 22:28
  • 3
    $\begingroup$ @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 $\endgroup$
    – El-Guest
    Commented Sep 8, 2018 at 22:33
  • 4
    $\begingroup$ 8!!!! is a multi-factorial $\endgroup$
    – JMP
    Commented Sep 8, 2018 at 22:33
13
$\begingroup$

$$8^2 + \sqrt{\frac{0!}{.\bar{1}}} = 64 + \sqrt{\frac{1}{\frac{1}{9}}} = 64 + 3 = 67$$ Note that $.\bar{1}$ is 0.111111... recurring.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ 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 $\endgroup$
    – Mr Pie
    Commented Sep 9, 2018 at 5:21
  • 1
    $\begingroup$ Nice answer! Welcome to Puzzling. $\endgroup$
    – Vicky
    Commented Sep 9, 2018 at 8:06
10
$\begingroup$

$8!!! - 12 - 0! = 8 \times 5 \times 2 - 12 - 1 = 80-13 = 67$

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ that is brilliant, but is your function after the 8 allowed??? $\endgroup$
    – tom
    Commented Sep 8, 2018 at 22:41
  • $\begingroup$ don't know. I guess we'll find out $\endgroup$
    – Marius
    Commented Sep 8, 2018 at 22:42
  • $\begingroup$ plus one and we will see :-) didn't know that was possible... $\endgroup$
    – tom
    Commented Sep 8, 2018 at 22:42
2
$\begingroup$

I base my answer on

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

Improve this answer
$\endgroup$
1
$\begingroup$

(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

Improve this answer
$\endgroup$
0
$\begingroup$

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.

Improve this answer
$\endgroup$
5
  • 1
    $\begingroup$ Please make use of the spoiler tag >! $\endgroup$
    – TheSimpliFire
    Commented Sep 9, 2018 at 13:37
  • 1
    $\begingroup$ Added spoiler tag. $\endgroup$
    – Arc
    Commented Sep 9, 2018 at 14:19
  • 1
    $\begingroup$ Why the downvotes? $\endgroup$
    – Arc
    Commented Sep 9, 2018 at 21:25
  • $\begingroup$ I did not downvote. $\endgroup$
    – TheSimpliFire
    Commented Sep 10, 2018 at 6:19
  • $\begingroup$ +1 Why downvote?? $\endgroup$
    – user44188
    Commented Sep 10, 2018 at 8:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.