Difference between revisions of "2014 AMC 10A Problems/Problem 21"
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Which means that <cmath>-\dfrac{5}{a} = -\dfrac{b}{3} \implies ab = 15</cmath> | Which means that <cmath>-\dfrac{5}{a} = -\dfrac{b}{3} \implies ab = 15</cmath> | ||
The only possible pairs <math>(a,b)</math> then are <math>(a,b) = (1,15), (3,5), (5,3), (15, 1)</math>. These pairs give respective <math>x</math>-values of <math>-5, -\dfrac{5}{3}, -1, -\dfrac{1}{3}</math> which have a sum of <math>\boxed{\textbf{(E)} \: -8}</math>. | The only possible pairs <math>(a,b)</math> then are <math>(a,b) = (1,15), (3,5), (5,3), (15, 1)</math>. These pairs give respective <math>x</math>-values of <math>-5, -\dfrac{5}{3}, -1, -\dfrac{1}{3}</math> which have a sum of <math>\boxed{\textbf{(E)} \: -8}</math>. | ||
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==See Also== | ==See Also== |
Revision as of 19:34, 23 November 2014
Problem
Positive integers and are such that the graphs of and intersect the -axis at the same point. What is the sum of all possible -coordinates of these points of intersection?
Solution
Note that when , the values of the equations should be equal by the problem statement. We have that Which means that The only possible pairs then are . These pairs give respective -values of which have a sum of .
See Also
2014 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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.