Five different positive integers are such that if we take any two of them, possibly the same number twice, exactly nine different sums may be obtained. Find the largest positive integer which can divide the sum of any five such numbers.
I have no idea how to approach this problem hints, suggestions, and solutions would all be appreciated.
Taken from the 2015 CIMC
Best Answer
Suppose the positive integers are $a<b<c<d<e$.
Then $2a<a+b<2b<b+c<2c<c+d<2d<d+e<2e$ give nine distinct sums. So these must be the only sums of pairs.
Then for $a+c$ we have $a+b<a+c<b+c$
So $a+c = 2b$
Similar reasoning will show that
$b+d=2c$;
$c+e=2d$;
$a+d=b+c$;
$b+e=c+d$; and
$a+e=2c$.
From all this we obtain: $(c-b)-(b-a)=(a+c-2b)=0$
So $c-b=b-a$.
Similarly we will end up with $e-d=d-c=c-b=b-a$.
So the five numbers are in arithmetic progression. So their sum is divisible by $5$ (the sum will be $5$ times the middle value).