A 2016 Problem

Is it possible for both the sum and product of 2016 integers to be 2016?

#NumberTheory

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
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`\boxed{123}`	$\boxed{123}$

Comments

Answer is yes. Here are the numbers:

$\{ \underbrace{ -1, -1, -1, \ldots, -1}_{\text{504 times}} , \underbrace{ 1, 1, 1, \ldots, 1}_{\text{1510 times}} , 2, 1008 \}$

The trick is to have an even number of (-1)'s so that the product is positive, then we construct a system of equations of the number of (-1)'s and (1)'s that satisfy the conditions.

Pi Han Goh - 5 years, 3 months ago

Is there like a more "mathematical" solution to it? More like a variable-type solution?

Like, I agree the trick works, but ......

Mehul Arora - 5 years, 3 months ago

What do you mean? Are you saying you want me to show my working?

Pi Han Goh - 5 years, 3 months ago

@Pi Han Goh – Very Precisely, yes. I mean, don't get me wrong, but how did you exactly wind up on this solution?

Mehul Arora - 5 years, 3 months ago

@Mehul Arora – Okay for starters. Let one of the numbers be 2016. So the product of the remaining numbers must be 1, and the sum of these 2015 remaining numbers is equal to 0, and these numbers must be $\pm \; 1$ only. Let $E$ and $P$ denote the number of $-1$ 's and $1$ 's, then we have $E + P = 2015$ and $P-E = 0$ . Solving these diophantine equation shows that there is no solution.

Now let two of the numbers be 2 and $\frac{2016}2 = 1008$ . So the product of the remaining numbers must be 1, and the sum of these 2014 remaining numbers is equal to 0,and these numbers must be $\pm \; 1$ only. Let $E$ and $P$ denote the number of $-1$ 's and $1$ 's, then we have $E + P = 2014$ and $P-E = 2016-2-1008$ . Solving these diophantine equation shows $E = 504, P = 1510$ , hence my answer above.

Note that there are more than one answer. Here's another solution:

$\{ -6, \underbrace{-1,-1,-1, \ldots , -1}_{\text{9 times}} ,\underbrace{1,1,1, \ldots , 1}_{\text{2002 times}} , 2, 2,4, 21 \}$

Pi Han Goh - 5 years, 3 months ago

@Pi Han Goh – Oh okay, so basically we show that one solution exists, and that ends the question. Hmm, great :)

Thanks for helping :D

Mehul Arora - 5 years, 3 months ago

Came up with the same set. :)

Kshitij Alwadhi - 5 years, 3 months ago

Hmmm. Are the integers distinct?

Aditya Kumar - 5 years, 3 months ago

No, that would've been mentioned. :)

Mehul Arora - 5 years, 3 months ago

Well, 2016 doesn't have 2016 distinct factors!

Ameya Daigavane - 5 years, 2 months ago

Ahha, certainly. Well noticed.

Mehul Arora - 5 years, 2 months ago

@Kunal Jain We have the solution.

Mehul Arora - 5 years, 3 months ago

I already knew it the 1 I wrote was 504, (-4), 1 X 1765, (-1) X 249

Kunal Jain - 5 years, 3 months ago

Where was this question asked?

Kshitij Alwadhi - 5 years, 3 months ago

This question was asked at some competition< id on't exactly remember the name.

Mehul Arora - 5 years, 3 months ago

@Mehul Arora – Ya, I saw the result.

Kshitij Alwadhi - 5 years, 3 months ago

@Kshitij Alwadhi – "Also, Bro I came third!'. I think by this he meant, coming third doesnt need congratulatios as the paper was super easy :P

Kunal Jain - 5 years, 3 months ago

@Kunal Jain – Oh ! So it was in that sense xD

Kshitij Alwadhi - 5 years, 3 months ago

And btw, Congratulations to u both and Vaibhav for NSTSE.

How was the paper this time?

Kshitij Alwadhi - 5 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$