# Difference between revisions of "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

Let the line be $y=mx+c$, with $c$ being a positive integer. Then we have $3=4m+c$ so $m=(3-c)/4$. Now the $x$-intercept is $-c/m = -4c/(3-c) = 4c/(c-3) = (4c-12+12)/(c-3) = 4+12/(c-3).$

We need the $x$-intercept to be positive, so $4c/(c-3)>0$ and $c>0$ together imply $c-3>0 \implies c>3$, so if the $x$-intercept is to be an integer, $(c-3)$ must be a positive factor of $12$. Hence we can get $4+1$, $4+2$, $4+3$, $4+4$, $4+6$, and $4+12$, and the only ones of these that are prime are $4+1=5$ and $4+3=7$.

The first case gives $c-3=12 \implies c=15$ and the second case gives $c-3=4 \implies c=7$, and both of these satisfy all the conditions of the problem, so the number of solutions is $\boxed{\textbf{(C) } 2}$.