Difference between revisions of "2009 AMC 10A Problems/Problem 16"
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</math> | </math> | ||
− | === Solution 1 | + | === Solution 1 = |
From <math>|a-b|=2</math> we get that <math>a=b\pm 2</math> | From <math>|a-b|=2</math> we get that <math>a=b\pm 2</math> | ||
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<math>-4-3-2=\boxed{-9}</math> | <math>-4-3-2=\boxed{-9}</math> | ||
− | 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 === | === Solution 2 === |
Revision as of 11:15, 11 June 2019
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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