Given any set$$A =\left\{ a_1, a_2, a_3, a_4 \right\}$$of four distinct positive integers, we denote the sum$$a_1+a_2+a_3+a_4$$by$$s_A$$Let$$n_A$$A denote the number of pairs (i, j) with $$1 ≤ i

Question

Given any set$$A =\left\{  a_1, a_2, a_3, a_4 \right\}$$of four distinct positive integers, we denote the sum$$a_1+a_2+a_3+a_4$$by$$s_A$$Let$$n_A$$A denote the number of pairs (i, j) with $$1 ≤ i < j ≤ 4$$for which$$a_i+a_j $$divides$$s_A$$Find all sets A of four distinct positive integers which achieve the largest possible value
of$$n_A$$

asnaseer · Answer

Here are my initial thoughts on this.
I /think/ the condition can be re-expressed as:$$a_i+a_j+a_k+a_l=m(a_i+a_j)$$where i, j, k and l are all unique positive integers between 1 and 4 inclusive, and m is some positive integer.
this leads to the condition:$$1+\frac{a_k+a_l}{a_i+a_j}=m\quad\implies a_k+a_l=n(a_i+a_j)$$where n is some other positive integer. Now, we need to pick pairs (i, j) and there are 6 ways of picking 2 out of 4 which leads to:$$\begin{align}
a_1+a_2&=n_1(a_3+a_4)\tag{1}\
a_1+a_3&=n_2(a_2+a_4)\tag{2}\
a_1+a_4&=n_3(a_2+a_3)\tag{3}\
a_2+a_3&=n_4(a_1+a_4)\tag{4}\
a_2+a_4&=n_5(a_1+a_3)\tag{5}\
a_3+a_4&=n_6(a_1+a_2)\tag{6}\
\end{align}$$Now, since $n_i$ are supposed to be positive integers, then only one of each of these pairs of equations must be true:$$
\begin{align}
(1)\text{ and }(6)\quad\text{ as }n_1=1/n_6\
(2)\text{ and }(5)\quad\text{ as }n_2=1/n_5\
(3)\text{ and }(4)\quad\text{ as }n_3=1/n_4
\end{align}$$We therefore only have 3 possible pairings:$$\begin{align}
a_1+a_2&=n_1(a_3+a_4)\tag{1}\
a_1+a_3&=n_2(a_2+a_4)\tag{2}\
a_1+a_4&=n_3(a_2+a_3)\tag{3}
\end{align}$$So the largest that $n_A$ can be is 3.

Have I made any mistakes in my reasoning so far?

turingtest · Answer

Not as far as I can see, but these are from the IMO questions, so I can't do them. I or someone else will get back to you on that to check your progress, because I just don't know if you're on the right track. 

I anticipated this group would require a team effort and much time to solve these problems, so let's try to recruit others who know what they're doing to help solve these.

Congrads on the progress though, looks like you're better than me. If you would like I can link you the answers, since they are online. Or would you prefer to keep trying yourself and wait for others to help us (this group was created about 3 hours ago, so give it some time before we get the heavy hitters in on this)?

Easier problems are up higher, more like the one you solved in the general math group. Have a gander at those if you want something more easy...

asnaseer · Answer

what is IMO?

turingtest · Answer

an international competition for math students. Here's the original file: http://assets.openstudy.com/updates/attachments/4ede4311e4b05ed8401a8546-ishaan94-1323190287159-2011_eng.pdf

asnaseer · Answer

oh - I originally thought you meant In My Opinion when you said IMO in the other maths group :-)
I would rather try and solve this myself and only look at  the answers as a very last resort.

turingtest · Answer

I like that attitude. I'll keep on them myself, but it will take some time for sure...

asnaseer · Answer

np - every now and then it's good to keep the brain busy with something it can't do that easily :-)

turingtest · Answer

kudos to that! need to flex the brain muscles now and then :)
I'm pretty sure James or Zarkon will show up here eventually and help us, but James had trouble with one already, so these are NOT easy!
Good luck!

asnaseer · Answer

thanks - good luck to all!