# 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, 2016 at 8:05
• @Артур I made a meta post related to actions on this question - meta.math.stackexchange.com/questions/22319/… Jan 1, 2016 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, 2016 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, 2016 at 11:44
• @Start, "protected" doesn't mean "protected against downvoting and/or closure," it means "protected against stupid answers." Jan 2, 2016 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, 2016 at 11:31
• By the way you solution is a nice one! Jan 1, 2016 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, 2016 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, 2016 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, 2016 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, 2016 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, 2016 at 10:22
• @quid Fair enough. I suppose we'll need a clarification from Peter Woolfitt Jan 2, 2016 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, 2016 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, 2016 at 5:43
• This is incorrect. $2016\ne 2^{22/2}-22^{2^2}$.
– JRN
Jan 2, 2016 at 15:40
• @JoelReyesNoche How about $\;\;2016 = 2^{22/2}-2^{2^2}\;2\;\;$ then.
– dxiv
Jan 2, 2016 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, 2016 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, 2016 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, 2016 at 16:14
• Yes, indeed I do- comment retyped-thanks. Jan 5, 2016 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, 2016 at 12:44
• This seems to be so far the shortest representation found (multiplication is missing in the 2nd formula of @paw88789 ). Jan 5, 2016 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, 2016 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, 2016 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, 2016 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, 2016 at 11:51
• Actually, using your $!!$ idea, the above can of course be written $$2016=\frac{(5+5)!}{5!5!!}$$ Jan 5, 2016 at 11:55
• @PeterWoolfitt That one is much more concise at 12 points. I like it. Jan 5, 2016 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 $8^{5/3}=32$)

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