Difference between revisions of "2014 AMC 10A Problems/Problem 21"
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==Solution 2== | ==Solution 2== | ||
− | + | Using part of what we know from Solution 1, notice that the value of <math>x</math> cannot be less than <math>-5</math>. We also know for the first equation that the values of <math>x</math> have to be <math>5</math> divided by something. Also, for the second equation, the values of <math>x</math> can only be <math>-\frac13,-\frac23,-\frac33, \dots</math>. Therefore, we see that, the only values common between the two sequences are <math>-1, -5, -\frac13,-\frac53</math>, and adding them up, we get for our answer, <math>\boxed{\textbf{(E)} \: -8}</math>. | |
==Video Solution by Pi Academy== | ==Video Solution by Pi Academy== |
Revision as of 16:33, 2 November 2024
Contents
[hide]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 1
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 .
Solution 2
Using part of what we know from Solution 1, notice that the value of cannot be less than . We also know for the first equation that the values of have to be divided by something. Also, for the second equation, the values of can only be . Therefore, we see that, the only values common between the two sequences are , and adding them up, we get for our answer, .
Video Solution by Pi Academy
https://youtu.be/uUxcMZVFSR8?si=zQWytkiPOVw1CRTG ~ Pi Academy
Video Solution by Richard Rusczyk
https://www.youtube.com/watch?v=-vZKwIazT08&list=PLyhPcpM8aMvKEM8u4Q-7Gi0rU5WU4WOb1&index=1 - AMBRIGGS
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.