2014 AMC 10A Problems/Problem 21

Revision as of 22:21, 6 April 2018 by Maheshbabuchapati (talk | contribs) (Solution)

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 $\boxed{\textbf{(E)} \: -8}$.


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 $\boxed{\textbf{(E)} \: -8}$.

See Also

2014 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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All AMC 10 Problems and Solutions

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