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Difference between revisions of "2017 AIME II Problems/Problem 5"
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We can substitute this value into the earlier equation to find that <math>x+y=3(607)-1030=1821-1030=\boxed{791}</math>.
 
We can substitute this value into the earlier equation to find that <math>x+y=3(607)-1030=1821-1030=\boxed{791}</math>.
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==Solution 3==
 +
Note that if <math>a > b > c > d</math> are the elements of the set, then <math>a+b > a+c > b+c, a+d, > b+d > c+d</math>. Thus we can assign <math>a+b = x, a+c = y, b+c = 320, a+d = 287, b+d =234, c+d=180</math>. Then <math>x  + y= (a+b) + (a+c) = 2((a+d)+(b+c))-((c+d)+(b+d)) = 791</math>.
  
 
=See Also=
 
=See Also=
 
{{AIME box|year=2017|n=II|num-b=4|num-a=6}}
 
{{AIME box|year=2017|n=II|num-b=4|num-a=6}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 10:57, 17 August 2017

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

Solution 3

Note that if $a > b > c > d$ are the elements of the set, then $a+b > a+c > b+c, a+d, > b+d > c+d$. Thus we can assign $a+b = x, a+c = y, b+c = 320, a+d = 287, b+d =234, c+d=180$. Then $x + y= (a+b) + (a+c) = 2((a+d)+(b+c))-((c+d)+(b+d)) = 791$.

See Also

2017 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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