Problem

A set contains four numbers. The six pairwise sums of distinct elements of the set, in no particular order, are $189$ , $320$ , $287$ , $234$ , $x$ , and $y$ . Find the greatest possible value of $x+y$ .

Solution 1

Let these four numbers be $a$ , $b$ , $c$ , and $d$ , where $a>b>c>d$ . $x+y$ needs to be maximized, so let $x=a+b$ and $y=a+c$ because these are the two largest pairwise sums. Now $x+y=2a+b+c$ needs to be maximized. Notice $2a+b+c=3(a+b+c+d)-(a+2b+2c+3d)=3((a+c)+(b+d))-((a+d)+(b+c)+(b+d)+(c+d))$ . No matter how the numbers $189$ , $320$ , $287$ , and $234$ are assigned to the values $a+d$ , $b+c$ , $b+d$ , and $c+d$ , the sum $(a+d)+(b+c)+(b+d)+(c+d)$ will always be $189+320+287+234$ . Therefore we need to maximize $3((a+c)+(b+d))-(189+320+287+234)$ . The maximum value of $(a+c)+(b+d)$ is achieved when we let $a+c$ and $b+d$ be $320$ and $287$ because these are the two largest pairwise sums besides $x$ and $y$ . Therefore, the maximum possible value of $x+y=3(320+287)-(189+320+287+234)=\boxed{791}$ .

Solution 2

Let the four numbers be $a$ , $b$ , $c$ , and $d$ , in no particular order. Adding the pairwise sums, we have $3a+3b+3c+3d=1030+x+y$ , so $x+y=3(a+b+c+d)-1030$ . Since we want to maximize $x+y$ , we must maximize $a+b+c+d$ .

Of the four sums whose values we know, there must be two sums that add to $a+b+c+d$ . To maximize this value, we choose the highest pairwise sums, $320$ and $287$ . Therefore, $a+b+c+d=320+287=607$ .

We can hi my nam this value into the earlier equation to find that $x+y=3(607)-1030=1821-1030=\boxed{791}$ .

2017 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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2017 AIME II Problems/Problem 5

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Problem

Solution 1

Solution 2

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