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
·
1 year, 6 months ago

@Pi Han Goh
–
Is there like a more "mathematical" solution to it? More like a variable-type solution?

Like, I agree the trick works, but ......
–
Mehul Arora
·
1 year, 6 months ago

Log in to reply

@Mehul Arora
–
What do you mean? Are you saying you want me to show my working?
–
Pi Han Goh
·
1 year, 6 months ago

Log in to reply

@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
·
1 year, 6 months ago

Log in to reply

@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
·
1 year, 6 months ago

Log in to reply

@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
·
1 year, 6 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
·
1 year, 6 months ago

## 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. – Pi Han Goh · 1 year, 6 months ago

Log in to reply

– Kshitij Alwadhi · 1 year, 6 months ago

Came up with the same set. :)Log in to reply

Like, I agree the trick works, but ...... – Mehul Arora · 1 year, 6 months ago

Log in to reply

– Pi Han Goh · 1 year, 6 months ago

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

– Mehul Arora · 1 year, 6 months ago

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

\[ \{ -6, \underbrace{-1,-1,-1, \ldots , -1}_{\text{9 times}} ,\underbrace{1,1,1, \ldots , 1}_{\text{2002 times}} , 2, 2,4, 21 \} \] – Pi Han Goh · 1 year, 6 months ago

Log in to reply

Thanks for helping :D – Mehul Arora · 1 year, 6 months ago

Log in to reply

@Kunal Jain We have the solution. – Mehul Arora · 1 year, 6 months ago

Log in to reply

How was the paper this time? – Kshitij Alwadhi · 1 year, 6 months ago

Log in to reply

– Kshitij Alwadhi · 1 year, 6 months ago

Where was this question asked?Log in to reply

– Mehul Arora · 1 year, 6 months ago

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

– Kshitij Alwadhi · 1 year, 6 months ago

Ya, I saw the result.Log in to reply

– Kunal Jain · 1 year, 6 months ago

"Also, Bro I came third!'. I think by this he meant, coming third doesnt need congratulatios as the paper was super easy :PLog in to reply

– Kshitij Alwadhi · 1 year, 6 months ago

Oh ! So it was in that sense xDLog in to reply

– Kunal Jain · 1 year, 6 months ago

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

Hmmm. Are the integers distinct? – Aditya Kumar · 1 year, 6 months ago

Log in to reply

– Ameya Daigavane · 1 year, 6 months ago

Well, 2016 doesn't have 2016 distinct factors!Log in to reply

– Mehul Arora · 1 year, 6 months ago

Ahha, certainly. Well noticed.Log in to reply

– Mehul Arora · 1 year, 6 months ago

No, that would've been mentioned. :)Log in to reply