29
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Add the four basic operators $\times\div+\,\;-$ and optionally brackets to:

$10 \quad 9 \quad 8 \quad 7 \quad 6 \quad 5 \quad 4 \quad 3 \quad 2 \quad 1$

To get the total $2016$.

Rules:

  • We are looking for the simplest solution - i.e. the least amount of characters (ignoring spaces). Please include your character count in your answer.
  • Keep the order; do not add or combine numbers.
  • Use all four operators at least once.

Credit for initial concept: Alex Bellos

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  • 4
    $\begingroup$ Is it one of each operator? And can you combine numbers (eg. 2 and 1 makes 21)? $\endgroup$ – JonTheMon Jan 4 '16 at 17:34
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    $\begingroup$ If there are multiple ways of doing this, do you want the most complex, the most simple, or some other criteria? $\endgroup$ – Aggie Kidd Jan 4 '16 at 17:38
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    $\begingroup$ Do the numbers need to be in that order in the equation? $\endgroup$ – JonTheMon Jan 4 '16 at 17:58
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    $\begingroup$ Fun fact, if Carat was allowed: 10 x 9 + 8 + 7 * 6 + 5 ^ 4 x 3 + 2 -1 (credit: @TheDanWoods Twitter) $\endgroup$ – rybo111 Jan 5 '16 at 0:20
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    $\begingroup$ @rybo111 there are actually 2 ways with a single carat: 10 x 9 + 8 + 7 x 6 + 5 ^ 4 x 3 + 2 - 1 10 x 9 + 8 x 7 - 6 + 5 ^ 4 x 3 + 2 - 1 edit: oops 2 not 5 - foiled by integer division! $\endgroup$ – ejrb Jan 5 '16 at 15:40

13 Answers 13

24
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22 characters

$10 \times 9 \times 8 \times 7 \times 6 \div 5 \div (4 - 3 + 2) \times 1$

I looked at @Will's answer and found a way to improve on it.

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  • 2
    $\begingroup$ @rybo111 - Brute force turns up no solutions without groupings using only +, -, *, and /. Closest two are: 10*9*8*7/6/5*4*3-2+1 = 2015.0 and 10*9*8*7/6/5*4*3+2-1 = 2017.0. This should be accepted. :) $\endgroup$ – Will Jan 4 '16 at 19:31
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    $\begingroup$ @Will keep that second one saved for next year! $\endgroup$ – Joel Rondeau Jan 4 '16 at 19:32
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    $\begingroup$ Couldn't this save one character? Seems to me the final operator is unnecessary. $\endgroup$ – Jordan Jan 4 '16 at 19:43
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    $\begingroup$ @rybo111 brackets are an implicit multiplication. The answer is exactly the same if the final * is removed, but with one less character. $\endgroup$ – Jordan Jan 4 '16 at 19:54
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    $\begingroup$ @rybo111 There is plenty of different mathematical notation that is accepted by mathematical professionals that doesn't work on Google, for various reasons (ranging from the fact that it can't be written as plain text, to that Google simply doesn't implement it). $\endgroup$ – nanofarad Jan 5 '16 at 0:39
19
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22 characters

I don't think you can beat Joel's answer at 22 characters, but there are some nice ways to tie it (including a variation of Joel's for completeness):

$(10 - 9 + 8 \times 7 \times 6 - 5 + 4) \times 3 \times 2 \times 1$
$10 - 9 + 8 \times 7 \times (6 \times 5 + 4 \times 3 \div 2) - 1$
$10 - 9 + 8 \times 7 \times (6 + 5 \times 4 \times 3 \div 2) - 1$
$10 - 9 + 8 \times 7 \times 6 \div (5 - 4) \times 3 \times 2 - 1$
$10 - 9 + 8 \times 7 \times 6 \times (5 + 4) \div 3 \times 2 - 1$
$10 - 9 + 8 \times 7 \times 6 \times (5 + 4 + 3) \div 2 - 1$
$10 + (9 \times 8 \times 7 - 6 + 5) \times 4 - 3 \times 2 \div 1$
$10 \times 9 \times 8 \times 7 \div (6 - 5 + 4 - 3 \div 2 - 1)$
$10 \times 9 \times 8 \times 7 \div (6 \times 5 \div 4 - 3 \times 2 + 1)$
$10 \times 9 \times 8 \times 7 \times 6 \div (5 \times 4 - 3 \times 2 + 1)$
$10 \times 9 \times 8 \times 7 \times 6 \div (5 + 4 \times 3 - 2 \times 1)$
$10 \times 9 \times 8 \times 7 \times 6 \div 5 \div (4 - 3 + 2) \times 1$

There are other ways but many of them are trivial (change $\div 1$ to $\times 1$ or vice-versa, or $\div 1)$ to $)\div 1$

If we didn't have the restriction that we need to use all the different operators there is also a very nice solution:

$ 10 \times 9 \times 8 \times 7 \times 6 \div (5 + 4 + 3 + 2 + 1)$

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  • 3
    $\begingroup$ The non-restriction version is so clean! $\endgroup$ – rybo111 Jan 5 '16 at 0:09
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    $\begingroup$ @rybo111 Yeah, I like how the / is right in the middle. $\endgroup$ – Paul Jan 5 '16 at 1:04
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    $\begingroup$ You found so many ways. Is there any generalized method to come with these? or just random guesses? $\endgroup$ – Maha Jan 5 '16 at 13:42
  • $\begingroup$ 22 characters is indeed the best possible solution as it cannot be done without parentheses (verified by exhaustive search) $\endgroup$ – ejrb Jan 5 '16 at 16:22
  • $\begingroup$ I don't understand why nobody is taking the hints in multiple comments. You can take many of these to 21 characters by simply removing any multiplication operators next to a parentheses. $\endgroup$ – Jordan Jan 6 '16 at 16:11
12
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I came up with:

$(10 - 9) \times 8 \times 7 \times 6 \times (5 - 4 + 3 + 2 \div 1)$

This is 9 operators and 2 required groupings, for a total of 24 characters.

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  • $\begingroup$ Updated question $\endgroup$ – rybo111 Jan 4 '16 at 18:17
  • $\begingroup$ All right, editing in division real quick $\endgroup$ – Will Jan 4 '16 at 18:18
  • $\begingroup$ Great stuff! Can anyone beat 24 characters? $\endgroup$ – rybo111 Jan 4 '16 at 18:23
  • $\begingroup$ I was just writing an answer with your edit. $\endgroup$ – tfitzger Jan 4 '16 at 18:45
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    $\begingroup$ You could also remove the two asterisks right next to the parentheses i.e. 6(5-4) = 6*(5-4) $\endgroup$ – Marco Bonelli Jan 4 '16 at 19:27
8
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$10 + 9 \times 8 - 7 + 654 \times 3 - 21$

17 Characters... Is mushing numbers together allowed? Also, yay Mathematica.

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  • $\begingroup$ Oh, wait, it isn't allowed. $\endgroup$ – Shane Di Dona Jan 5 '16 at 0:32
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    $\begingroup$ Still another cool solution though! $\endgroup$ – rybo111 Jan 5 '16 at 0:36
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    $\begingroup$ Also missing $ ÷ $ $\endgroup$ – Jamie Barker Jan 6 '16 at 8:06
  • $\begingroup$ I like your solution and your use of the word mushing! How do you find something like this using Mathematica? Is there a special command you used? $\endgroup$ – CJD Jan 6 '16 at 13:11
6
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If we allow for implicit multiplication of parenthesized expressions then the following solutions, all of length 20, become possible

$10\times 9\times 8\times 7(6\div 5-4+3) 2\times 1$
$10\times 9\times 8\times 7 (6\div 5-4+3) 2\div 1$
$10\times 9\times 8 (7-6\div 5-4+3-2) 1$

This list has been generated via exhaustive search, and excludes needlessly parenthesizing expressions that only contain multiplication or division.

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  • $\begingroup$ Bravo! 21 was the previous leader with implicit multiplication $\endgroup$ – rybo111 Jan 6 '16 at 22:35
3
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21 chars

$10 \times 9 \times 8 (7 - 6 \div 5 - 4 + 3 - 2 \times 1)$

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1
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22 Chars:

10 x 9 x 8 x 7 x 6/(5 + 4 + 3 + 2 + 1)

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  • 1
    $\begingroup$ @PaulPro already posted this solution $\endgroup$ – rybo111 Jan 5 '16 at 8:51
1
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22 chars

±10*9*8*7*6/(5+4+3+2+1)

If ± is valid as usage of - character, then you get two answers, which of one is correct :)

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1
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Here is one more with 22 characters not mentioned in @PaulPro's answer:

10*9*8*7*6/(5*4-3-2*1)

Edit: As @DanHenderson pointed out, this has no + operator.

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  • $\begingroup$ Has no + operator. $\endgroup$ – Dan Henderson Jan 5 '16 at 14:45
1
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24 chars:

$(10-9+8) \times 7 \times (6 \times 5 + 4-3 + 2-1)$

$9 \times 7 \times 32 = 2016$

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0
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22 20 characters

10 + 9 * 8 * 7 * 4 - 6 / 3 - 5 - 2 - 1

Without order :(

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  • 3
    $\begingroup$ I count 20 characters. $\endgroup$ – Crispy Jan 4 '16 at 20:11
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    $\begingroup$ You have changed the order of the numbers so this answer is invalid but otherwise well done $\endgroup$ – RedLaser Jan 4 '16 at 21:18
0
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21!

10*9*8*7*6/(5+4+3+2+1)

But not valid, I guess...

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  • 1
    $\begingroup$ This was already included in @PaulPro’s answer, and a couple of others’. $\endgroup$ – Peter LeFanu Lumsdaine Jan 5 '16 at 13:53
0
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27 characters (missing /):

(10-9)(8*7)(6-5)(4)(3)(2+1)

and 22 characters (missing -):

10*9*8*7*6/(5+4+3+2+1)

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  • $\begingroup$ Missing minus in last suggestion, and missing / in first $\endgroup$ – Viktor Mellgren Jan 5 '16 at 10:33

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