# Computing $2016$ using basic operations on the fewest integers, in sequence

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.

• Can you use unary minus to negate numbers? Jan 13, 2016 at 1:03
• No you cannot negate numbers in any way Jan 13, 2016 at 1:03
• I am asking for the least number $n$ using any operator to get 2016. Jan 13, 2016 at 1:06
• $exp$ is "to the power of"? Jan 13, 2016 at 1:08
• Jan 13, 2016 at 1:29

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

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

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.

Here is a 7-number solution:

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