Difference between revisions of "2009 AMC 10A Problems/Problem 16"
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Therefore, the only possible values of <math>|a-d|</math> are 9, 5, 3, and 1. Their sum is <math>\boxed{18}</math>. | Therefore, the only possible values of <math>|a-d|</math> are 9, 5, 3, and 1. Their sum is <math>\boxed{18}</math>. | ||
− | + | == Solution 2 == | |
If we add the same constant to all of <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math>, we will not change any of the differences. Hence we can assume that <math>a=0</math>. | If we add the same constant to all of <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math>, we will not change any of the differences. Hence we can assume that <math>a=0</math>. |
Revision as of 15:46, 22 August 2019
Contents
Problem
Let ,
,
, and
be real numbers with
,
, and
. What is the sum of all possible values of
?
Solution 1
From we get that
Similarly, and
.
Substitution gives . This gives
. There are
possibilities for the value of
:
,
,
,
,
,
,
,
Therefore, the only possible values of are 9, 5, 3, and 1. Their sum is
.
Solution 2
If we add the same constant to all of ,
,
, and
, we will not change any of the differences. Hence we can assume that
.
From we get that
, hence
.
If we multiply all four numbers by , we will not change any of the differences. Hence we can WLOG assume that
.
From we get that
.
From we get that
.
Hence , and the sum of possible values is
.
See Also
2009 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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