2014 AMC 10A Problems/Problem 21: Difference between revisions
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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>. | ||
It is curious that all of these ordered pairs result in the same line when plugged into the two equations. As an exercise, ask yourself why that is. | |||
==Solution 2== | ==Solution 2== | ||
Revision as of 21:46, 11 June 2018
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
![\[0 = ax + 5 \implies x = -\dfrac{5}{a}\]](http://latex.artofproblemsolving.com/9/b/5/9b5974079f962de0b0cdffd7cc3e068e6aa9753a.png)
![\[0 = 3x+b \implies x= -\dfrac{b}{3}\]](http://latex.artofproblemsolving.com/0/5/d/05ddc94c34a7ca9558c5927b090cbe03dbd143a6.png)
Which means that ![\[-\dfrac{5}{a} = -\dfrac{b}{3} \implies ab = 15\]](http://latex.artofproblemsolving.com/9/a/8/9a816e2e78ddb95d2446fe65f792d4d65d844642.png)
The only possible pairs 
then are 
. These pairs give respective -values of 
which have a sum of 
.
It is curious that all of these ordered pairs result in the same line when plugged into the two equations. As an exercise, ask yourself why that is.
Solution 2
First, notice that the value of x cannot exceed 5 because the minimum value for a is 1. Also, notice that for the second equation, it intersects x at 
and so on. We then realize that the only integer values for x are 
and 
. We also see that for a fraction to be the value of x, the numerator must divide 5 evenly. So, the only other values are 
and 
. Adding, we get .
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination
