Difference between revisions of "2009 AMC 10A Problems/Problem 16"
Coltsfan10 (talk | contribs) |
|||
Line 54: | Line 54: | ||
Hence <math>|a-d|=|d|\in\{1,3,5,9\}</math>, and the sum of possible values is <math>1+3+5+9 = \boxed{18}</math>. | Hence <math>|a-d|=|d|\in\{1,3,5,9\}</math>, and the sum of possible values is <math>1+3+5+9 = \boxed{18}</math>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/z_W-Z9CHPR4 | ||
+ | |||
+ | ~savannahsolver | ||
== See Also == | == See Also == |
Revision as of 20:58, 19 December 2020
Contents
[hide]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. (This is due to the fact that we are calculating |d| at the end ~Williamgolly) Hence we can WLOG assume that .
From we get that .
From we get that .
Hence , and the sum of possible values is .
Video Solution
~savannahsolver
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.