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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15
  • 4
    $\begingroup$ Is it one of each operator? And can you combine numbers (eg. 2 and 1 makes 21)? $\endgroup$
    – JonTheMon
    Jan 4, 2016 at 17:34
  • 2
    $\begingroup$ 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, 2016 at 17:38
  • 3
    $\begingroup$ Do the numbers need to be in that order in the equation? $\endgroup$
    – JonTheMon
    Jan 4, 2016 at 17:58
  • 1
    $\begingroup$ 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, 2016 at 0:20
  • 1
    $\begingroup$ @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, 2016 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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11
  • 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, 2016 at 19:31
  • 11
    $\begingroup$ @Will keep that second one saved for next year! $\endgroup$ Jan 4, 2016 at 19:32
  • 1
    $\begingroup$ Couldn't this save one character? Seems to me the final operator is unnecessary. $\endgroup$
    – Jordan
    Jan 4, 2016 at 19:43
  • 1
    $\begingroup$ @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, 2016 at 19:54
  • 3
    $\begingroup$ @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, 2016 at 0:39
18
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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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6
  • 3
    $\begingroup$ The non-restriction version is so clean! $\endgroup$
    – rybo111
    Jan 5, 2016 at 0:09
  • 2
    $\begingroup$ @rybo111 Yeah, I like how the / is right in the middle. $\endgroup$
    – Paul
    Jan 5, 2016 at 1:04
  • 4
    $\begingroup$ You found so many ways. Is there any generalized method to come with these? or just random guesses? $\endgroup$
    – Maha
    Jan 5, 2016 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, 2016 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, 2016 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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6
  • $\begingroup$ Updated question $\endgroup$
    – rybo111
    Jan 4, 2016 at 18:17
  • $\begingroup$ All right, editing in division real quick $\endgroup$
    – Will
    Jan 4, 2016 at 18:18
  • $\begingroup$ Great stuff! Can anyone beat 24 characters? $\endgroup$
    – rybo111
    Jan 4, 2016 at 18:23
  • $\begingroup$ I was just writing an answer with your edit. $\endgroup$
    – tfitzger
    Jan 4, 2016 at 18:45
  • 1
    $\begingroup$ You could also remove the two asterisks right next to the parentheses i.e. 6(5-4) = 6*(5-4) $\endgroup$ Jan 4, 2016 at 19:27
8
$\begingroup$

$10 + 9 \times 8 - 7 + 654 \times 3 - 21$

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

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4
  • $\begingroup$ Oh, wait, it isn't allowed. $\endgroup$ Jan 5, 2016 at 0:32
  • 4
    $\begingroup$ Still another cool solution though! $\endgroup$
    – rybo111
    Jan 5, 2016 at 0:36
  • 1
    $\begingroup$ Also missing $ ÷ $ $\endgroup$ Jan 6, 2016 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, 2016 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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1
  • $\begingroup$ Bravo! 21 was the previous leader with implicit multiplication $\endgroup$
    – rybo111
    Jan 6, 2016 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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2
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
  • 1
    $\begingroup$ @PaulPro already posted this solution $\endgroup$
    – rybo111
    Jan 5, 2016 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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1
  • $\begingroup$ Has no + operator. $\endgroup$ Jan 5, 2016 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
$\begingroup$

22 20 characters

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

Without order :(

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2
  • 3
    $\begingroup$ I count 20 characters. $\endgroup$
    – Crispy
    Jan 4, 2016 at 20:11
  • 2
    $\begingroup$ You have changed the order of the numbers so this answer is invalid but otherwise well done $\endgroup$
    – RedLaser
    Jan 4, 2016 at 21:18
0
$\begingroup$

21!

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

But not valid, I guess...

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1
  • 1
    $\begingroup$ This was already included in @PaulPro’s answer, and a couple of others’. $\endgroup$ Jan 5, 2016 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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1
  • $\begingroup$ Missing minus in last suggestion, and missing / in first $\endgroup$ Jan 5, 2016 at 10:33

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