Difference between revisions of "2009 AMC 10A Problems/Problem 16"
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== Solution == | == Solution == | ||
+ | === Solution 1 === | ||
− | + | From <math>|a-b|=2</math> we get that <math>a=b\pm 2</math> | |
− | + | Similarly, <math>b=c\pm3</math> and <math>c=d\pm4</math>. | |
− | + | Substitution gives <math>a=d\pm 4\pm 3\pm 2</math>. This gives <math>|a-d|=|\pm 4\pm 3\pm 2|</math>. There are <math>2^3=8</math> possibilities for the value of <math>\pm 4\pm 3\pm2</math>: | |
− | + | <math>4+3+2=\boxed{9}</math>, | |
− | |||
<math>4+3-2=\boxed{5}</math>, | <math>4+3-2=\boxed{5}</math>, | ||
+ | |||
<math>4-3+2=\boxed{3}</math>, | <math>4-3+2=\boxed{3}</math>, | ||
+ | |||
<math>-4+3+2=\boxed{1}</math>, | <math>-4+3+2=\boxed{1}</math>, | ||
+ | |||
<math>4-3-2=\boxed{-1}</math>, | <math>4-3-2=\boxed{-1}</math>, | ||
+ | |||
<math>-4+3-2=\boxed{-3}</math>, | <math>-4+3-2=\boxed{-3}</math>, | ||
+ | |||
<math>-4-3+2=\boxed{-5}</math>, | <math>-4-3+2=\boxed{-5}</math>, | ||
+ | |||
<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 === |
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 01:44, 8 February 2010
Contents
[hide]Problem
Let , , , and be real numbers with , , and . What is the sum of all possible values of ?
Solution
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 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 |