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Previous question: Use two 2's, two 1's, and two 8's to make the number 2018

Background: I was trying to add some new elements (hopefully creative and interesting) to this traditional type of number-making puzzles, but was clearly beaten by some smart brains which solved the puzzle beautifully without using the intended element. So I am going to make another attempt. I will be happy to see the puzzle solved by traditional ways as well (although @Oray has commented that it is impossible), in which case I am beaten again but I can learn something one way or the other. (I actually have learnt a lot from the answers, thank you all)

Question: I know the year 2016 has passed ... but I am going to publish this riddle anyway because I cannot think of a more elegant one right now. Rules:

  1. Use exactly four 8's in the equation, no more, no less.
  2. Allowed symbols: $+$, $-$, $\times$, $\div$, $($, $)$, $\sqrt{\quad}$, $!$. Arbitrary functions (such as the logarithm) are not allowed.
  3. It is OK to use numbers as superscript (exponent or the power for the radical symbol).
  4. Concatenation is allowed although using $8(8\div8)$ to construct $81$ is not allowed.
  5. Ceiling or flooring is not allowed. $88\div(8+8)=5.5$, not $5$ or $6$.
  6. The use of decimal point or scientific notation is not allowed.
  7. The final solution must be an equation. using $!$ to make $!=$ ("not equal to" in some programming languages) is not allowed
  8. $+$ or $-$ alone as superscript has different meanings in different contexts. It is not allowed here. Superscript can only be numbers.
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    $\begingroup$ I checked all possibilities with a program code, no answer for 2016. $\endgroup$
    – Oray
    Jan 17, 2018 at 8:55
  • $\begingroup$ @Oray Thank you for your confirmation. That's why I say I want people to jump out of the box (I may just as well have added the lateral-thinking tag to the question). $\endgroup$
    – Weijun Zhou
    Jan 17, 2018 at 8:57
  • $\begingroup$ I did not find a solution for this with four 8's, but incidentally, I found an alternative solution with the conditions of the previous riddle (two 1's, 2's and 8's). I just don't know where to post it. $\endgroup$
    – Thern
    Jan 17, 2018 at 11:36
  • $\begingroup$ @Nebr You can post it on the other question and refer to the comments here, if you wish, so that people won't think you have mistaken 2018 for 2016. $\endgroup$
    – Weijun Zhou
    Jan 17, 2018 at 12:15
  • $\begingroup$ Yeah, but there it would be misfitting as well. I think it's maybe easiest to do it directly here: The year 2016 is (at least partly) the year 5776 in the Jewish calendar, which is $(8-1)!+((2+1)!)!+8\times 2$. $\endgroup$
    – Thern
    Jan 17, 2018 at 12:35

4 Answers 4

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Obscure enough? If this is what you have in mind, then I also know your intended solution for the other problem :).

$\displaystyle\binom{8\times8}{\sqrt{\sqrt{8+8}}} = \binom{64}{2} = 2016$

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  • $\begingroup$ That's great! What do you think the intended answer for the other problem is? $\endgroup$
    – Weijun Zhou
    Jan 17, 2018 at 12:16
  • $\begingroup$ Check my edit over there and you'll see. $\endgroup$
    – slvrbld
    Jan 17, 2018 at 12:18
  • $\begingroup$ Yes that's a nice solution, but actually when I wrote down that problem for the first time I was thinking about double factorial :). $\endgroup$
    – Weijun Zhou
    Jan 17, 2018 at 12:29
  • $\begingroup$ Fair enough. That's even more obscure than what would possibly cross my mind ;). $\endgroup$
    – slvrbld
    Jan 17, 2018 at 12:32
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    $\begingroup$ Isn't the binomial an arbitrary function you're not allowed to use? $\endgroup$
    – Yanko
    Jan 17, 2018 at 13:55
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Not sure if that's the part of lateral-thinking or not

8! / ( ( 8 + 8 ) / 8 ) = 20160
So we've got 2016 plus 0, but what difference does it make? :)

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  • $\begingroup$ Thank you for your interest. It is Interesting, but this is not the intended answer. The intended one is really making the number 2016, not a number that is of the pattern *2016*. $\endgroup$
    – Weijun Zhou
    Jan 17, 2018 at 9:12
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$$8!/((8+8)/.8))$$
$$8!/(16/.8)$$
$$8!/(20)$$
$$40320/20$$
$$2016$$

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    $\begingroup$ Nice try, but note point 6. $\endgroup$
    – Weijun Zhou
    Jan 17, 2018 at 9:17
  • $\begingroup$ :) my bad! almost there $\endgroup$
    – Leonardo Peña
    Jan 17, 2018 at 9:23
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Do you mean any number can be used as superscript in point 3? If so,

8!/((8+8^1/3)*(8^1/3))

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    $\begingroup$ I don't mean that, I just mean you can place the number (formed by the existing digits) as superscript. $\endgroup$
    – Weijun Zhou
    Jan 18, 2018 at 15:19

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