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.

@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:

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestAnswer 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.

Log in to reply

Came up with the same set. :)

Log in to reply

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

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

Log in to reply

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

Log in to reply

Log in to reply

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:

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

Log in to reply

Thanks for helping :D

Log in to reply

@Kunal Jain We have the solution.

Log in to reply

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

How was the paper this time?

Log in to reply

Where was this question asked?

Log in to reply

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

Log in to reply

Log in to reply

Log in to reply

Log in to reply

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

Log in to reply

Hmmm. Are the integers distinct?

Log in to reply

Well, 2016 doesn't have 2016 distinct factors!

Log in to reply

Ahha, certainly. Well noticed.

Log in to reply

No, that would've been mentioned. :)

Log in to reply