2014 AMC 10A Problems/Problem 21

Problem

Positive integers $a$ and $b$ are such that the graphs of $y=ax+5$ and $y=3x+b$ intersect the $x$ -axis at the same point. What is the sum of all possible $x$ -coordinates of these points of intersection?

$\textbf{(A)}\ {-20}\qquad\textbf{(B)}\ {-18}\qquad\textbf{(C)}\ {-15}\qquad\textbf{(D)}\ {-12}\qquad\textbf{(E)}\ {-8}$

Solution 1

Note that when $y=0$ , the $x$ values of the equations should be equal by the problem statement. We have that $0 = ax + 5 \implies x = -\dfrac{5}{a}$ $0 = 3x+b \implies x= -\dfrac{b}{3}$ Which means that $-\dfrac{5}{a} = -\dfrac{b}{3} \implies ab = 15$ The only possible pairs $(a,b)$ then are $(a,b) = (1,15), (3,5), (5,3), (15, 1)$ . These pairs give respective $x$ -values of $-5, -\dfrac{5}{3}, -1, -\dfrac{1}{3}$ which have a sum of $\boxed{\textbf{(E)} \: -8}$ .

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 $0, -\dfrac{1}{3}, -\dfrac{2}{3}, -1$ and so on. We then realize that the only integer values for x are $-1$ and $-5$ . 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 $-\dfrac{5}{3}$ and $-\dfrac{1}{3}$ . Adding, we get $\boxed{\textbf{(E)} \: -8}$ .

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.

Page

Toolbox

Search

2014 AMC 10A Problems/Problem 21

Contents

Problem

Solution 1

Solution 2

See Also