8
$\begingroup$

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.
$\endgroup$
  • 1
    $\begingroup$ I checked all possibilities with a program code, no answer for 2016. $\endgroup$ – Oray Jan 17 '18 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 '18 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 '18 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 '18 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 '18 at 12:35
13
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ That's great! What do you think the intended answer for the other problem is? $\endgroup$ – Weijun Zhou Jan 17 '18 at 12:16
  • $\begingroup$ Check my edit over there and you'll see. $\endgroup$ – slvrbld Jan 17 '18 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 '18 at 12:29
  • $\begingroup$ Fair enough. That's even more obscure than what would possibly cross my mind ;). $\endgroup$ – slvrbld Jan 17 '18 at 12:32
  • 4
    $\begingroup$ Isn't the binomial an arbitrary function you're not allowed to use? $\endgroup$ – Yanko Jan 17 '18 at 13:55
0
$\begingroup$

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? :)

$\endgroup$
  • $\begingroup$ 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 '18 at 9:12
0
$\begingroup$

$$8!/((8+8)/.8))$$
$$8!/(16/.8)$$
$$8!/(20)$$
$$40320/20$$
$$2016$$

$\endgroup$
  • 1
    $\begingroup$ Nice try, but note point 6. $\endgroup$ – Weijun Zhou Jan 17 '18 at 9:17
  • $\begingroup$ :) my bad! almost there $\endgroup$ – Leonardo Peña Jan 17 '18 at 9:23
0
$\begingroup$

Do you mean any number can be used as superscript in point 3? If so,

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

$\endgroup$
  • 2
    $\begingroup$ 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 '18 at 15:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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