Difference between revisions of "1994 AHSME Problems/Problem 28"

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<math> \textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4</math>
 
<math> \textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4</math>
 
==Solution==
 
==Solution==
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Let the line be <math>y=mx+c</math>, with <math>c</math> being a positive integer. Then we have <math>3=4m+c</math> so <math>m=(3-c)/4</math>. Now the <math>x</math>-intercept is <math>-c/m = -4c/(3-c) = 4c/(c-3) = (4c-12+12)/(c-3) = 4+12/(c-3).</math>
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We need the <math>x</math>-intercept to be positive, so <math>4c/(c-3)>0</math> and <math>c>0</math> together imply <math>c-3>0 \implies c>3</math>, so if the <math>x</math>-intercept is to be an integer, <math>(c-3)</math> must be a positive factor of <math>12</math>. Hence we can get <math>4+1</math>, <math>4+2</math>, <math>4+3</math>, <math>4+4</math>, <math>4+6</math>, and <math>4+12</math>, and the only ones of these that are prime are <math>4+1=5</math> and <math>4+3=7</math>.
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The first case gives <math>c-3=12 \implies c=15</math> and the second case gives <math>c-3=4 \implies c=7</math>, and both of these satisfy all the conditions of the problem, so the number of solutions is <math>\boxed{\textbf{(C) } 2}</math>.

Revision as of 03:36, 18 February 2018

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 $\boxed{\textbf{(C) } 2}$.

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