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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?

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    $\begingroup$ "Best" - shortest, more concise. "Nerdiest" - using mathematical functions in interesting ways. $\endgroup$ Mar 7, 2014 at 18:18
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    $\begingroup$ Or $2kg_{27}$ . $\endgroup$
    – user88595
    Mar 7, 2014 at 18:25
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    $\begingroup$ $13^3-13^2-13^1-13^0$ (Courtesy of the OEIS) $\endgroup$
    – Jack M
    Mar 7, 2014 at 18:41
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    $\begingroup$ @JackM Go ahead and make that an answer! $\endgroup$ Mar 7, 2014 at 18:42
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    $\begingroup$ Did you know that 2014 is the only natural number that precedes 2015 and succeeds 2013? $\endgroup$
    – evil999man
    Mar 8, 2014 at 18:04

15 Answers 15

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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.

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    $\begingroup$ Use the keyword:easy to get simple results like this. $\endgroup$
    – Ypnypn
    Mar 9, 2014 at 19:25
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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.

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    $\begingroup$ 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. $\endgroup$
    – Mr Lister
    Mar 8, 2014 at 10:51
  • $\begingroup$ @MrLister thank you very much for the comment $\endgroup$ Mar 8, 2014 at 13:46
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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

$$2014=5^5-1111$$

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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 $17$." $:)$

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

SSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS$0$

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    $\begingroup$ I like it ;) although I have a feeling that writing this out on a board could be rather monotonous $\endgroup$ Mar 7, 2014 at 18:26
  • $\begingroup$ How is this $2014$? $\endgroup$
    – K. Rmth
    Mar 7, 2014 at 19:51
  • $\begingroup$ @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) $\endgroup$
    – Tyler
    Mar 7, 2014 at 19:59
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    $\begingroup$ @K.Rmth $S$ is the successor function. So, for example, $S(4) = 5$. In general, $S(x) = x + 1$. $\endgroup$ Mar 7, 2014 at 20:00
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    $\begingroup$ 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.) $\endgroup$
    – Unwisdom
    Mar 7, 2014 at 20:23
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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".

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  • $\begingroup$ You definitely have the right idea here! I am intrigued as to how you worked out the first two in such a short time $\endgroup$ Mar 7, 2014 at 18:23
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    $\begingroup$ @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. $\endgroup$
    – FireGarden
    Mar 7, 2014 at 19:00
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    $\begingroup$ This answer misses the point. These identities don't reveal anything interesting about the number 2014. They would work for any number. $\endgroup$
    – user85798
    Mar 10, 2014 at 5:29
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How about

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

or, if you like

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

Alternatively:

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

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

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    $\begingroup$ Using "2014" in the expression itself is surely cheating. $\endgroup$ Mar 7, 2014 at 22:45
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$$ 2014 = 2^{2\times2^2 \times (2\times2^2-2)}-2\times (2\times2)^2-2 $$

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

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

10

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Maybe do:

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

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    $\begingroup$ aka 2^11 - 34, since 2^11 = 2048 $\endgroup$
    – smci
    Mar 8, 2014 at 18:06
  • $\begingroup$ I like my way better :-) $\endgroup$
    – Patrick
    Mar 8, 2014 at 21:44
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    $\begingroup$ I was just pointing out why they're equivalent... $\endgroup$
    – smci
    Mar 9, 2014 at 15:04
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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..

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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.

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

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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 $17$.

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