Small Representations of $2016$

Question

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. — Alex Fok, Jan 01 '16 at 08:05
@Артур I made a meta post related to actions on this question - http://meta.math.stackexchange.com/questions/22319/appropriate-use-of-moderator-power-with-respect-to-question-closure — Peter Woolfitt, Jan 01 '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. — achille hui, Jan 02 '16 at 02:03
@PeterWoolfitt The questions have been raised: Are square roots allowable? Binomial coefficients? — paw88789, Jan 02 '16 at 10:44
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. — Start wearing purple, Jan 02 '16 at 11:44
@paw88789 When I originally wrote the question, I did not intend to allow such solutions. However, I am indeed still interested in them. — Peter Woolfitt, Jan 02 '16 at 13:55
@Start, "protected" doesn't mean "protected against downvoting and/or closure," it means "protected against stupid answers." — Gerry Myerson, Jan 02 '16 at 15:20
I have edited to provide a more precise question, and clarify the allowed actions. — Peter Woolfitt, Jan 02 '16 at 16:23
Then I guess $;;aa_{55};;$ doesn't qualify (where $a$ is digit $36$ in base $55$ using the common $;0123..9ABC..Zabc...;$ notation). — dxiv, Jan 02 '16 at 18:47
The American Mathematical Monthly: $2016=3^2+4^2+5^2+6^2+7^2+8^2+9^2$. See https://www.facebook.com/AmerMathMonthly/posts/955542504494734 (some other expressions for 2016 are given in the comments.) — Martin Sleziak, Jan 05 '16 at 08:00
@MartinSleziak, as noted by Gerry, there seems to be an error in your exponents. — JRN, Jan 12 '16 at 01:08
@JoelReyesNoche Sorry, I did not notice that. But the main point of the comment was to add the link, which seems to work. — Martin Sleziak, Jan 12 '16 at 01:10

Start wearing purple · Answer 1 · 2016-01-01T19:01:34.077

19

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 $64$.

edited Jan 01 '16 at 19:01

answered Jan 01 '16 at 11:17

Start wearing purple

54,260

2

Why don't you use $\binom{2^{2^2+2}}{2}$? This way you use one less symbol! – Alex Fok Jan 01 '16 at 11:31
By the way you solution is a nice one! – Alex Fok Jan 01 '16 at 11:31
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)." This does not appear to be a valid answer, as "binomial coefficient" does not appear in this list. – quid Jan 02 '16 at 10:19
2

@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. – Start wearing purple Jan 02 '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. – epimorphic Jan 04 '16 at 23:22

score 17 · Answer 2 · answered Jan 02 '16 at 04:42

17

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

answered Jan 02 '16 at 04:42

Gerry Myerson

185,413

1

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! – Paolo Franchi Jan 05 '16 at 12:39

paw88789 · Answer 3 · 2016-01-01T06:15:12.517

10

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

Edit: Here's another nine:

$2016=3!(333+3)$

edited Jan 01 '16 at 06:15

answered Jan 01 '16 at 06:07

paw88789

41,907

1

"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 02 '16 at 10:22
1

@quid Fair enough. I suppose we'll need a clarification from Peter Woolfitt – paw88789 Jan 02 '16 at 10:43

dxiv · Answer 4 · 2016-01-03T19:24:00.703

6

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

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

edited Jan 03 '16 at 19:24

answered Jan 02 '16 at 01:11

dxiv

77,867

You missed `$\times$'. 11 symbols are needed. – Alex Fok Jan 02 '16 at 05:29
@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 02 '16 at 05:43
This is incorrect. $2016\ne 2^{22/2}-22^{2^2}$. – JRN Jan 02 '16 at 15:40
1

@JoelReyesNoche How about $;;2016 = 2^{22/2}-2^{2^2};2;;$ then. – dxiv Jan 02 '16 at 17:29
" I assumed that if an implicit × wasn't counted in other examples" well not really, this was at the cost of two parenthesis. – Surb Jan 03 '16 at 19:10
4

@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 03 '16 at 19:26
$$\frac{2\left(222+2\right)}{.\dot{2}}$$ Using 10 symbols and two little dots. Surely they don't count. :) – Ian Miller Jan 12 '16 at 13:59

score 6 · Answer 5 · answered Jan 05 '16 at 03:49

6

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. $(7 \times 7 - 7) \times (7 \times 7 - \frac77)$.

answered Jan 05 '16 at 03:49

Noam D. Elkies

26,275

1

You mean $2016 = 4 \left| {\rm SL}_2(2,8) \right|$, not $2014$. – Noam D. Elkies Jan 05 '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. – Geoff Robinson Jan 05 '16 at 16:14
Yes, indeed I do- comment retyped-thanks. – Geoff Robinson Jan 05 '16 at 16:15

Noam D. Elkies · Answer 6 · 2016-01-05T22:16:14.197

6

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.]

edited Jan 05 '16 at 22:16

answered Jan 05 '16 at 12:03

Noam D. Elkies

26,275

2

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: $$\frac{.8\times 8!}{8+8}$$ – Peter Woolfitt Jan 05 '16 at 12:44
1

This seems to be so far the shortest representation found (multiplication is missing in the 2nd formula of @paw88789 ). – Start wearing purple Jan 05 '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 $\sqrt9=3$. – Noam D. Elkies Jan 06 '16 at 17:11

score 5 · Answer 7 · answered Jan 05 '16 at 02:15

5

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

answered Jan 05 '16 at 02:15

Sandeep Silwal

8,363

1

Although pretty neat, he did say single digits. – Ahmed S. Attaalla Jan 05 '16 at 03:30

score 3 · Answer 8 · answered Jan 11 '16 at 20:07

3

Here's a devil of an answer:

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

answered Jan 11 '16 at 20:07

zz20s

6,852

Does the downvoter care to comment? – zz20s Jan 15 '16 at 04:06

score 1 · Answer 9 · answered Jan 05 '16 at 05:16

1

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.

answered Jan 05 '16 at 05:16

Axoren

2,353

I got another $5$ one - though the idea came from user Paulpro on the linked puzzling stackexchange question: $$2016=\frac{(5+5)!}{(5+5+5)5!}$$ – Peter Woolfitt Jan 05 '16 at 11:51
2

Actually, using your $!!$ idea, the above can of course be written $$2016=\frac{(5+5)!}{5!5!!}$$ – Peter Woolfitt Jan 05 '16 at 11:55
@PeterWoolfitt That one is much more concise at 12 points. I like it. – Axoren Jan 05 '16 at 17:33

score 0 · Answer 10 · answered Jan 05 '16 at 12:59

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 $8^{5/3}=32$)

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

Small Representations of $2016$

10 Answers10

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

Linked