This is another number formation question. You must use each of $2$, $1$ and $8$ exactly twice to make the number $2018$. The rules are

  1. Allowed symbols: $+$, $-$, $\times$, $\div$, $($, $)$, $\sqrt{\quad}$, $!$. Arbitrary functions (such as the logarithm) are not allowed.
  2. The ordering of the numbers are not important.
  3. It is OK to use numbers as superscript (exponent or the power for the radical symbol).
  4. Concatenation is allowed although using $(2+1)1$ to construct $31$ is not allowed.
  5. Ceiling or flooring is not allowed. $211\div2=105.5$, not $105$ or $106$.
  6. The use of decimal point or scientific notation is not allowed.

    For example, $$(2+8)\times(2+1)!\times\sqrt{1+8}=180$$ is a valid construction, although this is not a solution because it does not equal $2018$.

Edit (after slvrbld posted his answer): Try to make $2108$ and $8102$ as well if you can.

(And of course, the more ways the better).

Second Edit I am thinking whether I can make all 4-digit numbers composed of 2, 0, 1 and 8 (numbers beginning with 0 do not count). I have solutions for some but not all of them. Once I find solution for all of them or find a well defined subset for which I have solutions, I will update it and make it a formal challenge here.

  • $\begingroup$ Was that edit really supposed to say 2108, or should it just be 2018 ? $\endgroup$
    – JPhi1618
    Commented Jan 16, 2018 at 16:32
  • $\begingroup$ @JPhi1618 The original question is about 2018, which we already have an answer here. The extra challenges are 2108 and 8102. $\endgroup$ Commented Jan 16, 2018 at 16:34
  • $\begingroup$ Thanks for the clarification. Also, note that if you accept an answer, others are greatly discouraged from providing new answers. If you want more traffic, I'd take back the check mark and only accept after a day or so. $\endgroup$
    – JPhi1618
    Commented Jan 16, 2018 at 16:50
  • $\begingroup$ @JPhi1618 Thank you for your advice. I will take it. $\endgroup$ Commented Jan 16, 2018 at 16:51
  • 1
    $\begingroup$ @Lawrence No, I mean the order of the numbers on the left hand side of the equation does not matter (Some similar puzzles requests that the numbers be in a specific order, like you can only put symbols in 2 1 8 2 1 8=2018, but you cannot swap the order of the numbers. I mean my puzzle is not of this kind). Anyway, 2180 is part of the challenge in my "2nd Edit" $\endgroup$ Commented Jan 17, 2018 at 9:47

5 Answers 5


A possible solution:

$ 1\times2 + 8!/(12+8) $

Edit: more obscure approach:

$\displaystyle\binom{8\times8}{1\times1\times2} + 2 = \binom{64}{2} + 2 = 2018$

  • $\begingroup$ That's a nice solution although I was thinking about something more obscure ... $\endgroup$ Commented Jan 16, 2018 at 15:49
  • 5
    $\begingroup$ Why go for obscure if it can be done in a clean way? :) $\endgroup$
    – slvrbld
    Commented Jan 16, 2018 at 16:01
  • 1
    $\begingroup$ I agree with you, so I think your solution is better than my original one. $\endgroup$ Commented Jan 16, 2018 at 16:03
  • $\begingroup$ @WeijunZhou, feel free to post your own answer after waiting a little for others to answer. $\endgroup$
    – JPhi1618
    Commented Jan 16, 2018 at 16:05
  • $\begingroup$ @JPhi1618 Sure, but I am just checking whether there are ways to achieve 2108 without using my "obscure" method, so I have edited the question and will wait to see. $\endgroup$ Commented Jan 16, 2018 at 16:06

For 8102:

$ 8112-8-2 $

For 2108:

$ 2118-8-2 $

  • 2
    $\begingroup$ Most simplistic solution, great. $\endgroup$
    – Tweakimp
    Commented Jan 16, 2018 at 16:40
  • $\begingroup$ You are really quick I must say. $\endgroup$ Commented Jan 16, 2018 at 16:42

To build up on slvrblds solution by using only the pure numbers (without concatenation):

$2018 = 2 + 8! / (2 \times (8 + 1 + 1))$

Edit: Got another one with a multifactorial (for obscurity reasons):

$2018 = 1 + 1 + 8! \times 2 \times 2 \div 8!!!$

  • 1
    $\begingroup$ I love this one. $\endgroup$ Commented Jan 16, 2018 at 16:56
  • 1
    $\begingroup$ Yep, nice one. Good usage of the leftover one from my solution! $\endgroup$
    – slvrbld
    Commented Jan 16, 2018 at 17:27
  • $\begingroup$ The new solution is also beautiful, and includes the element from my intended answer. $\endgroup$ Commented Jan 17, 2018 at 12:51

First time here :)

$8! / (18+2) + 2*1$

  • $\begingroup$ Thank you for your effort, but usage of the extra 3 in the superscript is not allowed, and you have three(now four) 2's in the formula while there should only be two 2's. $\endgroup$ Commented Jan 16, 2018 at 16:54
  • $\begingroup$ Oh, wow I really didn't read that very thoroughly :P $\endgroup$
    – oyvind
    Commented Jan 16, 2018 at 16:55
  • 1
    $\begingroup$ Never mind, now this one is the same as slvrbld's but it's perfectly OK to keep it as it is. $\endgroup$ Commented Jan 16, 2018 at 17:00
  • 3
    $\begingroup$ Not completely exactly the same. It's 12+8 vs 18+2. ;) $\endgroup$
    – Thern
    Commented Jan 16, 2018 at 17:32

Are we restricted to Equations?

Argument: you need some sort of method to perceive the final result.

The only negative aspect of this answer is that it is not universal. Blind people would not understand how to read the digit of the final printed number as zero (0). Mathematics can be used by the blind, Braille is their chief source of information.

The symbol for 0 (zero) in Braille is a 3 x 2 matrix of:

| 0 1 | | 1 1 | | 0 0 |

which doesn't hold up mathematically as in Braille, using the sequence would result in 2818

(see Braille symbols here: https://www.pharmabraille.com/pharmaceutical-braille/the-braille-alphabet/)

Nevertheless, zero (0) is the absence of any sort of number, so processing this information (even for blind people reading Braille) this can allow for an intuitive understanding of the second digit as the non-integer zero (0 = chaos, absence of comprehensive entity).

Graphical solution using:

Sequence: 2 1 2 8 1 8 Keypress: 2 back 1 back 2 back 8 1 8

enter image description here

Lateral Thinking? You could convert to binary I suppose and manipulate them in that manner. Also, converting numbers to unicode and tweaking them would be an exercise in lateral thinking..

Anyway, I'm not a mathematician but after watching the movie "Arrival" two days ago, I see that skills in linguistics can help decipher 'classical' problems with left-field (right-hemisphere) tactics ;)

  • 2
    $\begingroup$ "Are we restricted to Equations?" Specifically you are restricted to use 6 numbers and the ways you could combine those 6 numbers were specified. Implicitly the standard meanings of the operators should be used. $\endgroup$ Commented Jan 17, 2018 at 14:09
  • $\begingroup$ Thank you for your long answer, it provides me some ideas that I can make use of in designing the next puzzles. However I think many of the ideas will be considered too lateral to be suitable for a good puzzle. $\endgroup$ Commented Jan 17, 2018 at 16:36
  • $\begingroup$ @WeijunZhou There is no limit to what makes a good puzzle. Nevertheless this puzzle did stoke some dying brain-cells back into life so I'm the one who should be saying thanks. Cheers. ;) $\endgroup$
    – NexusInk
    Commented Jan 18, 2018 at 3:02

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.