# 1994 AHSME Problems/Problem 28

## Problem

In the $xy$-plane, how many lines whose $x$-intercept is a positive prime number and whose $y$-intercept is a positive integer pass through the point $(4,3)$? $\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4$

## Solution 1

The line with $x$-intercept $a$ and $y$-intercept $b$ is given by the equation $\frac{x}{a} + \frac{y}{b} = 1$. We are told $(4,3)$ is on the line so $$\frac{4}{a} + \frac{3}{b} = 1 \implies ab - 4b - 3a = 0 \implies (a-4)(b-3)=12$$

Since $a$ and $b$ are integers, this equation holds only if $(a-4)$ is a factor of $12$. The factors are $1, 2, 3, 4, 6, 12$ which means $a$ must be one of $5, 6, 7, 8, 10, 16$. The only members of this list which are prime are $a=5$ and $a=7$, so the number of solutions is $\boxed{\textbf{(C) } 2}$.

## Solution 2

Let $C = (4,3)$, $DF=a$, and $AD=b$. As stated in the problem, the $x$-intercept $DF=a$ is a positive prime number, and the $y$-intercept $AD=b$ is a positive integer.

Through similar triangles, $\frac{AB}{BC}=\frac{CE}{EF}$, $\frac{b-3}{4}=\frac{3}{a-4}$, $(a-4)(b-3)=12$

The only cases where $a$ is prime are: $$\begin{cases} a-4=1 & a=5 \\ b-3=12 & b=15 \end{cases}$$ $$and$$ $$\begin{cases} a-4=3 & a=7 \\ b-3=4 & b=5 \end{cases}$$

So the number of solutions are $\boxed{\textbf{(C) }2}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 