Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In my class, I always write the date using mathematical formulas, or cool little equations. I want to show my students that even the most mundane seeming number often has fascinating features, and its own beauty - that's the reason I got into mathematics, and I want to pass it down.

For example, for 30, I wrote $\frac{6!}{4!}$.

Now, 2014 has kind of stumped me. The best I could come up with was this, using factorials:

$$(2!(2!+2!(4!))+3!)(4!-3!+1!)$$

But, this isn't the most attractive, or interesting, equation.

What do you think the best (nerdiest?) way to write 2014 is?

share|improve this question

closed as primarily opinion-based by mrf, TMM, Rahul, Austin Mohr, LTS Mar 10 at 5:22

Many good questions generate some degree of opinion based on expert experience, but answers to this question will tend to be almost entirely based on opinions, rather than facts, references, or specific expertise.If this question can be reworded to fit the rules in the help center, please edit the question.

2  
"Best" - shortest, more concise. "Nerdiest" - using mathematical functions in interesting ways. –  James Williams Mar 7 at 18:18
2  
The shortest and most concise way to write $2014$, I suspect, is to write $2014$. A nerdy twist? express in terms of its prime factors: $2\cdot 19\cdot 53$ –  amWhy Mar 7 at 18:19
5  
Or $2kg_{27}$ . –  user88595 Mar 7 at 18:25
52  
$13^3-13^2-13^1-13^0$ (Courtesy of the OEIS) –  Jack M Mar 7 at 18:41
3  
@JackM Go ahead and make that an answer! –  James Williams Mar 7 at 18:42

15 Answers 15

Looking $2014$ up in the OEIS turns up:

$$2014=13^3-13^2-13^1-13^0$$

In general, looking a number up in the OEIS is probably a reasonable way to turn up pleasing identities.

share|improve this answer
1  
Use the keyword:easy to get simple results like this. –  Ypnypn Mar 9 at 19:25

Give them the following :

$$(-2+0+1+4)^{(2+0+1+4)}-(2+0+1+4)^{(-2+0+1+4)}+(2+0-1+4)^{(-2+0+1+4)}+(2+0-1+4)\cdot(-2+0+1+4)^2=?$$

and tell them to compute the result.

All you can see is only 2014 with some sign changed and of course the result is simply
$$3^7-7^3+5^3+5\cdot3^2=2014$$
It will look better on a board.

share|improve this answer
1  
I like this one; With all due respect, but many of the other answers don't look more interesting to me than 2⋅10³+0⋅10²+1⋅10¹+4⋅10⁰. –  Mr Lister Mar 8 at 10:51
    
@MrLister thank you very much for the comment –  Konstantinos Gaitanas Mar 8 at 13:46

With all due credit to this base 13 answer on codegolf.SE:

$$2014=BBC_{13}$$


Or just playing with my calculator, I like the look of

$$2014=5^5-1111$$

share|improve this answer
2  
+1 for the second –  Ross Millikan Mar 10 at 2:31

I hope you will enjoy the following spoof :

$\qquad\quad$ I remember once going to see him for the Holidays, and remarked that the number of the upcoming year seemed to me rather a dull one, and that I hoped it was not an unfavorable omen. "No," he replied, "it is a very interesting number; it is the smallest number which can be expressed as the product of three distinct primes, which are congruent modulo $17$." $:)$

share|improve this answer

What it lacks in brevity, it makes up for in nerdiness:

SSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS$0$

share|improve this answer
4  
I like it ;) although I have a feeling that writing this out on a board could be rather monotonous –  James Williams Mar 7 at 18:26
    
How is this $2014$? –  K. Rmth Mar 7 at 19:51
    
@K.Rmth well, it appears to be 2014 S's and 1 zero. Not sure what the 0 is for (or why S was chosen) –  Tyler Mar 7 at 19:59
16  
@K.Rmth $S$ is the successor function. So, for example, $S(4) = 5$. In general, $S(x) = x + 1$. –  Gamma Function Mar 7 at 20:00
4  
Peano arithmetic has one constant symbol, 0, and one unary function symbol, $S$. Every natural number is either $0$ (in which case it is not the successor of anything) or the successor of some (unique) other natural number. Every natural number can thus be written as a finite number of applications of the successor function to the constant zero. Thus the natural numbers are defined recursively, which makes this essentially the minimal structure needed to enable inductive type proofs. (I'm simplifying slightly - some formulations of Peano Arithmetic permit nonstandard natural numbers.) –  Unwisdom Mar 7 at 20:23

Here are some cryptic ones: From the Gaussian integral we have $$2014 = \frac{4028}{\sqrt{\pi}}\int_{0}^{\infty}e^{-x^2}dx$$ and from the Basel Problem we have: $$2014 = \frac{12084}{\pi^2}\sum_{i=1}^{\infty}{\frac{1}{i^2}}$$ Here are some that (arguably) has deep meanings and roots: $$ 2014 = 2\cdot19\cdot53$$ $$2014 = 2^{11} - 34$$ For some trigonometry we have: $$2014 = \frac{4}{\cos^3{\frac{\pi}{9}}\cdot\cos^3{\frac{2\pi}{9}}\cdot\cos^3{\frac{4\pi}{9}}} - 34$$ It depends on perception, really. There are probably arguably infinitely many ways to write $2014$ in a "short, snappy, cool, and nerdy way".

share|improve this answer
    
You definitely have the right idea here! I am intrigued as to how you worked out the first two in such a short time –  James Williams Mar 7 at 18:23
1  
@JamesWilliams Well, you have 2*2014=4028 and 6*2014=12084, where the 2 and the 6 belonged to the (rather well known) cases where the 2014 is removed from both sides, and rearranged so the sum/integral are on their own side. –  FireGarden Mar 7 at 19:00
1  
This answer misses the point. These identities don't reveal anything interesting about the number 2014. They would work for any number. –  LTS Mar 10 at 5:29

How about

$$3\cdot6!-5!-4!-2!$$

or, if you like

$$(6!-5!)+(6!-4!)+(6!-2!)$$

Alternatively:

$$6!2!+4!4!-2!0!$$

share|improve this answer

(Self answering question) Find the integral part of the unique real root of the equation $$\log_2 x+\log_{20}x+\log_{201}x+\log_{2014}x -2-0-14+\frac{1}{20+\frac{1}{14}} = 0$$

share|improve this answer
6  
Using "2014" in the expression itself is surely cheating. –  David Richerby Mar 7 at 22:45

$$ 2014 = 2^{2\times2^2 \times (2\times2^2-2)}-2\times (2\times2)^2-2 $$

Equations like this can be made for any number, not just 2014.

share|improve this answer

Binary: 11111011110 Hexadecimal: 7DE

Image the students perplexing expression when they see: 1101/11/111111011110 or C/3/7DE Tell them to write this date in this form on there notes. Guaranteed they will show it to their friends or family.

Wow I can't believe I haven't thought of doing this with my students. As a rookie high school math teacher I am always looking for new 'hooks' with my students. Great idea and thank you!

Lesson planning using stack exchange? Who knew..

share|improve this answer

Maybe do:

$$2014 = \sum_{k=0}^{11}\binom{11}{k}-34$$

share|improve this answer
1  
aka 2^11 - 34, since 2^11 = 2048 –  smci Mar 8 at 18:06
    
I like my way better :-) –  Patrick Mar 8 at 21:44
1  
I was just pointing out why they're equivalent... –  smci Mar 9 at 15:04

Using a base 2014 number system, it would be expressed as:

10

share|improve this answer

How about multiple mathematical formulae (using only digits and simple operators) for every year (except 2102) for the rest of the century? e.g.,

$$2014 = 10*9*8*7/6/5*4*3-2*1$$

Read more about this in the fascinating Wolfram Blog. To generate these equations, take a look at the answers for this SO question.

share|improve this answer

If anyone went to HCSSiM, we have a tradition of worshiping the number $17$.

So I would say the coolest way to express $2014$ is $$2014=2 \times 19 \times 51$$

It is easy to show that this neat fact: $2014$ is in fact the smallest number that can be expressed as a product of three distinct positive integers that are all congruent modulo $17$.

share|improve this answer

I would express this as a subtraction of powers of two, i.e. $$ 2^{11} - 2^5 - 2^1 $$

share|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.