# Small Representations of $2016$

It's the new year at least in my timezone, and to welcome it in, I ask for small representations of the number $2016$.

Rules: Choose a single decimal digit ($1,2,\dots,9$), and use this chosen digit, combined with operations, to create an expression equal to $2016$. Each symbol counts as a point, and the goal is to minimize the total number of points.

Example (my best so far):

$$2016=\frac{(4+4)!}{4!-4}$$

This expression scores 11 points. That is 2 points for the parentheses, 4 points for the $4$s, 1 point for $+$, 2 points for the $!$s, 1 point for the fraction, and 1 point for the $-$.

Allowable actions: basic arithmetic operations (addition, subtraction, multiplication, division), exponentiation, factorials, repeated digits (i.e. if you are working with the digit $7$, you can use $77$ for 2 points), and use of parentheses.

What is the minimum number of points for an expression of the above form equaling 2016, and what are those minimum expressions?

Note that by "Use a single decimal digit" I mean you may only use one of the digits $1$ through $9$, so for example, you can't save in the above expression just by using $8!$ instead of $(4+4)!$ because you would still have the $4!-4$ part.

This question is mostly for fun, but could have some relevance to students who participate in thematic math competitions this year.

• 'There are either too many possible answers, or good answers would be too long for this format.' Why? I don't think that's the case for this interesting problem. Jan 1 '16 at 8:05
• @Артур I made a meta post related to actions on this question - meta.math.stackexchange.com/questions/22319/… Jan 1 '16 at 21:47
• +1 for the part for assigning concrete points to the operations. this make it have a chance to have an optimal solution. Jan 2 '16 at 2:03
• I don't understand why this question is ferociously downvoted and systematically closed. It is definitely more interesting then protected question about why Arnold considered American students stupid. Jan 2 '16 at 11:44
• @Start, "protected" doesn't mean "protected against downvoting and/or closure," it means "protected against stupid answers." Jan 2 '16 at 15:20

With binomial coefficients ($8$ symbols): $$2016={64 \choose 2}={2^{2^2+2} \choose 2}$$ It can be expected that many olympiad problems in $2016$ will use this combinatorial property.

P.S. Special thanks to Alex Fok for minus one symbol in $64$.

• Why don't you use $\binom{2^{2^2+2}}{2}$? This way you use one less symbol! Jan 1 '16 at 11:31
• By the way you solution is a nice one! Jan 1 '16 at 11:31
• "Allowable actions: basic arithmetic operations, exponentiation, factorials, and repeated digits (i.e. if you are working with the digit 7, you can use 77 for 2 points)." This does not appear to be a valid answer, as "binomial coefficient" does not appear in this list.
– quid
Jan 2 '16 at 10:19
• @quid Sure. However given the recreational context of the question I thought this may be an interesting remark. Maybe I should have made it a comment though. Jan 2 '16 at 11:41
• Since the question is now "What is the minimum number of points for an expression of the above form equaling 2016, and what are those minimum expressions?", this does not even attempt to answer the question. Jan 4 '16 at 23:22

$$\sum_3^{3\times3}n^3$$ is 7 points, right?

• I doubt about this being 7 points because I would require to write $n=3$ under the sum. Moreover, I interpret OP not to allow this ($n$ is neither a digit nor an operation symbol). Very nice though! Jan 5 '16 at 12:39

How about $2016=4\sqrt{4}(4^4-4)$ using $9$ symbols.

Edit: Here's another nine:

$2016=3!(333+3)$

• "Allowable actions: basic arithmetic operations, exponentiation, factorials, and repeated digits (i.e. if you are working with the digit 7, you can use 77 for 2 points)." It is perhaps not exactly clear if taking a square root is considered as basic arithmetic operation, but I would say "no."
– quid
Jan 2 '16 at 10:22
• @quid Fair enough. I suppose we'll need a clarification from Peter Woolfitt Jan 2 '16 at 10:43

$2016 = 2^{\,22\,/\,2} - 2\;2^{\,2^{\,2}}\;$ with just elementary operations (10 symbols).

[EDIT]  P.S.  A few more variations with and without the × disputed in the comments. $$2016 = 2^{\,22\,/\,2} - 2^{\,2^{\,2}}\;2 = 2^{\,22\,/\,2} - 2 \cdot 2^{\,2^{\,2}} = \sqrt 2 ^{\,22} - 2 \; 2^{\,2^{\,2}} = \sqrt 2 ^{\,22} - 2 \cdot 2^{\,2^{\,2}}$$

• You missed $\times$'. 11 symbols are needed. Jan 2 '16 at 5:29
• @AlexFok There is no × in the posted formula ;-) I assumed that if an implicit × wasn't counted in other examples, it wouldn't count here, either. But of course the rules are not entirely clearly cut as stated. Happy New 11111100000!
– dxiv
Jan 2 '16 at 5:43
• This is incorrect. $2016\ne 2^{22/2}-22^{2^2}$. Jan 2 '16 at 15:40
• @JoelReyesNoche How about $\;\;2016 = 2^{22/2}-2^{2^2}\;2\;\;$ then.
– dxiv
Jan 2 '16 at 17:29
• @Surb I still don't see why $2^{2^2}2$ would require a ×` when $4 \sqrt 4$ doesn't. But anyway, this was supposed to be all for fun, so I edited my answer to keep everybody happy.
– dxiv
Jan 3 '16 at 19:26

Since $2016 = \# {\rm GL}_2({\bf Z} / 7 {\bf Z})$ [alas I don't remember from whom I learned this a year or so ago], we can even use $7$'s nicely, though not enough to win this "contest", e.g. $(7 \times 7 - 7) \times (7 \times 7 - \frac77)$.

• You mean $2016 = 4 \left| {\rm SL}_2(2,8) \right|$, not $2014$. Jan 5 '16 at 13:25
• Curiously enough, we also have 2016=4|SL(2,8)| which allows an expression (even further from optimal) using only the digit 8 or one using only the digit 4. Jan 5 '16 at 16:14
• Yes, indeed I do- comment retyped-thanks. Jan 5 '16 at 16:15

9 symbols using, appropriately, the digit 9: $$\frac{9!}{99 + 9 \times 9}$$ (I thought $6.6 \times 6! - 6!$ was another 9-symbol solution if decimal points are allowed, but it actually gives $4032$ which is exactly twice too big.)

[added later: then there's the puzzle of writing 201 (not 2016) using four 9's and any number of the standard operations $+$ $-$ $\times$ $\div$ ! $\sqrt{\phantom0}$ and parentheses.]

• Neat answer! I was trying to find one like that. Using decimal points, it's actually possible to get another 9-symbol solution with the digit 8: $$\frac{.8\times 8!}{8+8}$$ Jan 5 '16 at 12:44
• This seems to be so far the shortest representation found (multiplication is missing in the 2nd formula of @paw88789 ). Jan 5 '16 at 13:32
• I see now that the $201$ puzzle has two kinds of solution. One, represented by $6^3 - 6 - 9$ (with $3=\sqrt9$ and $6=3!$), is related with another formula for $2016$ with five $9$'s, involving more symbols but no decimal concatenation: $9(6^3+9)-9$. But there's another, more exotic approach to $201$ that doesn't require the use of $\sqrt9=3$. Jan 6 '16 at 17:11

Make sure you follow your New Year's resolution $24$ hours a day, $7$ days a week, for $12$ months $24 \cdot 7 \cdot 12 = 2016$.

• Although pretty neat, he did say single digits. Jan 5 '16 at 3:30

Here's a devil of an answer:

$$2016=666+666+666+6+6+6$$

• Does the downvoter care to comment? Jan 15 '16 at 4:06

## The first $5$ answer (25 points, lousy)

$55 \frac {55} 5 \frac {5!!} 5 + 5 (55 - 5!!) + \frac 5 5$

Didn't see a $5$ answer yet. Threw my hand in the ring for some digit representation.

• I got another $5$ one - though the idea came from user Paulpro on the linked puzzling stackexchange question: $$2016=\frac{(5+5)!}{(5+5+5)5!}$$ Jan 5 '16 at 11:51
• Actually, using your $!!$ idea, the above can of course be written $$2016=\frac{(5+5)!}{5!5!!}$$ Jan 5 '16 at 11:55
• @PeterWoolfitt That one is much more concise at 12 points. I like it. Jan 5 '16 at 17:33

13 points using $2$s

$$2016 = 2^{2^2} ( 2^{2^2} 2^2 2 - 2)$$

19 points using $6$s

$$2016 = 6! + 6! + 6! - (6+6)(6+6)$$

Many points using $8$s (involving $8^{5/3}=32$)

$$2016 = \left( \frac{8 \cdot 8}{88 - 8 \cdot 8} - \frac{8}{8} \right) \left(8 \cdot 8 - \frac{8}{8} \right)$$