Difference between revisions of "2017 AIME II Problems/Problem 5"
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Of the four sums whose values we know, there must be two sums that add to <math>a+b+c+d</math>. To maximize this value, we choose the highest pairwise sums, <math>320</math> and <math>287</math>. Therefore, <math>a+b+c+d=320+287=607</math>. | Of the four sums whose values we know, there must be two sums that add to <math>a+b+c+d</math>. To maximize this value, we choose the highest pairwise sums, <math>320</math> and <math>287</math>. Therefore, <math>a+b+c+d=320+287=607</math>. | ||
− | We can | + | We can hi my nam this value into the earlier equation to find that <math>x+y=3(607)-1030=1821-1030=\boxed{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 16:40, 3 April 2017
Contents
[hide]Problem
A set contains four numbers. The six pairwise sums of distinct elements of the set, in no particular order, are ,
,
,
,
, and
. Find the greatest possible value of
.
Solution 1
Let these four numbers be ,
,
, and
, where
.
needs to be maximized, so let
and
because these are the two largest pairwise sums. Now
needs to be maximized. Notice
. No matter how the numbers
,
,
, and
are assigned to the values
,
,
, and
, the sum
will always be
. Therefore we need to maximize
. The maximum value of
is achieved when we let
and
be
and
because these are the two largest pairwise sums besides
and
. Therefore, the maximum possible value of
.
Solution 2
Let the four numbers be ,
,
, and
, in no particular order. Adding the pairwise sums, we have
, so
. Since we want to maximize
, we must maximize
.
Of the four sums whose values we know, there must be two sums that add to . To maximize this value, we choose the highest pairwise sums,
and
. Therefore,
.
We can hi my nam this value into the earlier equation to find that .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.