Let N be a positive integer. Two players, Yuri and Zeta, play the following game:
= Zeta picks a value for N;
= Yuri chooses N real numbers, not necessarily distinct, and keeps them secret;
= Yuri then writes all pairwise sums on a sheet of paper and gives it to Zeta (there are n(n-1)/2 such sums, not necessarily distinct);
= Zeta wins if she finds correctly the initial N numbers chosen by Yuri with only one guess.
Can Zeta be sure to win for the following cases?
a) N = 4 b) N = 5 c) N = 6
Justify your answer(s).
[For example, when N = 3, Yuri may choose the numbers 1, 5, 9, which have the pairwise sums of 6, 10 and 14 and can easily be resolved by simple algebra.]
Note, this has been adapted from the Junior Balkan MO 2013, Problem 4.
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For the case with three numbers, assume you have values x, y, and z arranged in ascending order. Then it follows that x + y is the smallest sum and y + z is the largest sum. Adding all pairwise sums gives 2(x+y+z), and dividing through by 2 and subtracting the smallest number on yuri's list
gives z, and similarly dividing by 2 and subtracting the largest number on yuri's list gives x.
For the case with four numbers, if you have x, y, and z you would have 3(w+x+y+z) and if you try to do the same thing, you get would still have two variables left over no matter what you chose. However trial and error shows that if you have numbers 1, 5, 7 and 9, the sums are now 6, 8, 10, 12, 14, and 16. These can be reproduced by using 2, 4, 6, and 10.
For the case with five numbers, you can again use the same way for three numbers. The sum would be 4(v + w + x + y + z), and if you subtract the biggest and smallest numbers simultaneously you can obtain x. You could obtain v by subtracting the next smallest sum in the list from x, and also get z by subtracting the second largest number. Then w and y could also be obtained by elimination.
For six numbers, I'd thm not certain. I know you can at least determine the sum of W+X by subtracting the largest and smallest values from the sum of all the pairwise sums divided by 5. If you start playing around with the numbers, you can get W+Y and V+X as well, which means you can also get U +Z. I'm sure that if you arrange the pairs more you could eventually come up with an answer, but it's late now and I want to sleep.
So my answers are:
N = 4 - No
N = 5 - Yes
N = 6 - Leaning towards NO, but I (literally) need to sleep on it.
N=6 takes a while to write out, but, say, starting from the smallest, we can easily figure out the first 3 numbers from the lowest 3 sums... the rest is algebra.
The actual test had the extra interesting case of N=8 :-)
hmmmm!!! This mathematics is a bit confusing sir. @rycharde.
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