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Using the operators $$+,-,\div,\times,\exp,(,),!$$ what is the least $n$ to come up with the number $2016$ using the sequence of numbers $1,2,3,\ldots,n$ in that order. You cannot combine numbers, so you cannot do $2~~3=23$ and you cannot negate values.

My solution consists of $10$ numbers. I want to see if someone can come up with the least use of operators in their answer.

Good luck! If no one is able to get less than $10$ numbers, I will post my answer.

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3 Answers 3

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$4$ number solution: $$\left((1+2)!\right)!+(3!)^4=2016$$

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  • $\begingroup$ Bravo.... surely unsurpassable. So how'd you find it? $\endgroup$ Jan 13, 2016 at 1:23
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    $\begingroup$ @DavidG.Stork Carefully $\endgroup$
    – user174622
    Jan 13, 2016 at 1:24
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    $\begingroup$ I was going to write a Mathematica program to search. But from now on I guess I should just be "careful" as that suffices! $\endgroup$ Jan 13, 2016 at 1:24
  • $\begingroup$ @DavidG.Stork If you don't mind, I would like to see the Mathematica program. $\endgroup$
    – user174622
    Jan 13, 2016 at 1:25
  • $\begingroup$ @DavidG.Stork There may be other variations of this solution, but I doubt there is anything less than $4$. $\endgroup$
    – user174622
    Jan 13, 2016 at 1:27
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Well, this is a 9 number solution.

$1-(2!\times 3! \times 4! \times (5-6) \times 7) +8-9=2016$

I have the feeling that this is not the smallest one, but this is the smallest I can find at the moment.

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Here is a 7-number solution:

$((1 + 2 * 3) * 4! - 5!) * 6 * 7$

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