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

Question

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.

I am asking for the least number $n$ using any operator to get 2016. — user304729, Jan 13 '16 at 01:06
So something like $((1 + 2) * 3 ^ {4!})$ would be valid if it equaled 2016, but not $1 ^ {-2}$. — shoover, Jan 13 '16 at 01:10
Related: http://puzzling.stackexchange.com/q/25039/5095 and http://math.stackexchange.com/q/1595709/18398 — JRN, Jan 13 '16 at 01:29
$2016$ is the smallest number of the form $2^{a+b}~\Big(2^{ab}-1\Big),$ with a and b distinct primes. — Lucian, Jan 13 '16 at 06:31

score 7 · Answer 1 · 2016-01-13T01:23:27.343

7

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

edited Jan 13 '16 at 01:23

answered Jan 13 '16 at 01:22

Bravo.... surely unsurpassable. So how'd you find it? – David G. Stork Jan 13 '16 at 01:23
1

@DavidG.Stork Carefully – Jan 13 '16 at 01:24
2

I was going to write a Mathematica program to search. But from now on I guess I should just be "careful" as that suffices! – David G. Stork Jan 13 '16 at 01:24
@DavidG.Stork If you don't mind, I would like to see the Mathematica program. – Jan 13 '16 at 01:25
@DavidG.Stork There may be other variations of this solution, but I doubt there is anything less than $4$. – Jan 13 '16 at 01:27
I just started to write it when your answer appeared, so I realized it wasn't worth my time. – David G. Stork Jan 13 '16 at 01:27
@DavidG.Stork Sounds fair – Jan 13 '16 at 01:28

score 2 · Answer 2 · answered Jan 13 '16 at 01:18

2

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.

answered Jan 13 '16 at 01:18

Chad Shin

2,160

score 2 · Answer 3 · answered Jan 13 '16 at 01:23

2

Here is a 7-number solution:

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

answered Jan 13 '16 at 01:23

shoover

595

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

3 Answers3