Difference between revisions of "2009 AMC 10A Problems/Problem 16"
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== Solution == | == Solution == | ||
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+ | Solution 1: | ||
+ | |||
+ | From <math>|a-b|=3</math> we get that <math>a=b+-2</math> | ||
+ | |||
+ | Similarly, <math>b=c+-3</math> and <math>c=d+-4</math>. | ||
+ | |||
+ | Substitution gives <math>a=d+-4+-3+-2</math>. This gives <math>|a-d|=|+-4+-3+-2|</math>. There are <math>2^3=8</math> possibilities for the value of <math>|+-4+-3+-2|</math>: | ||
+ | |||
+ | <math>4+3+2=\boxed{9}</math>, | ||
+ | <math>4+3-2=\boxed{5}</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{-3}</math>, | ||
+ | <math>-4-3+2=\boxed{-5}</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>. | ||
+ | |||
+ | 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 17:22, 7 February 2010
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 |