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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
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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

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## Comments

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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.

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

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

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

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$\pm \; 1$ 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.

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 beNow 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:

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

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Thanks for helping :D

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Came up with the same set. :)

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Hmmm. Are the integers distinct?

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No, that would've been mentioned. :)

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Well, 2016 doesn't have 2016 distinct factors!

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Ahha, certainly. Well noticed.

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@Kunal Jain We have the solution.

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

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Where was this question asked?

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

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

How was the paper this time?

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