Difference between revisions of "2008 UNCO Math Contest II Problems/Problem 2"

Problem

Let $S = \left \{a,b,c,d \right \}$ be a set of four positive integers. If pairs of distinct elements of $S$ are added, the following six sums are obtained: $5, 10, 11, 13, 14, 19.$ Determine the values of $a, b, c$, and $d.$ [Hint: there are two possibilities.]

Solution

The problem doesn't tell us that $a, b, c,$ and $d$ are distinct integers, but we can prove it easily. If we assume that these four positive integers are all distinct, then we will obtain six distinct pair-wise sums, since the number of ways to choose two objects out of four total where order does not matter is $(4 * 3)/2 = 6$. If any of the elements $a, b, c,$ or $d$ are equal to another, at least one of these distinct pair-wise sums would disappear, because at least two of the sums would be identical. However, we can see that the pair-wise sums we are given are all distinct, and that there are 6 of them, meaning that elements $a, b, c,$ and $d$ are all distinct.

Before we move on to considering cases, it would help to consider what information might help save time. If we assume that $a < b < c < d$, it is obvious that $a + b = 5$ and $c + d = 19$, as those are the minimum and maximum pair-wise sums. We can do better though, because we know that $a + c = 10$ and $b + d = 14$, since those are the next smallest and next largest pair-wise sums.

With this knowledge, we can start considering cases. If $a = 1$ and $b = 4$, we can use the equations we found to determine that $c = 9$ and $d = 10$. Similarly, if $a = 2$ and $b = 3$, $c = 8$ and $d = 11$. Both of these solutions are valid, and are also the only solutions, because the only two possibilities for $a$ and $b$ were exhausted.

Thus, the answer is $\{1,4,9,10\}$ or $\{2,3,8,11\}$