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

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Since <math>a</math> and <math>b</math> are integers, this equation holds only if <math>(a-4)</math> is a factor of <math>12</math>.  The factors are <math>1, 2, 3, 4, 6, 12</math> which means <math>a</math> must be one of <math>5, 6, 7, 8, 10, 16</math>.  The only members of this list which are prime are <math>a=5</math> and <math>a=6</math>, so the number of solutions is <math>\boxed{\textbf{(C) } 2}</math>.
 
Since <math>a</math> and <math>b</math> are integers, this equation holds only if <math>(a-4)</math> is a factor of <math>12</math>.  The factors are <math>1, 2, 3, 4, 6, 12</math> which means <math>a</math> must be one of <math>5, 6, 7, 8, 10, 16</math>.  The only members of this list which are prime are <math>a=5</math> and <math>a=6</math>, so the number of solutions is <math>\boxed{\textbf{(C) } 2}</math>.
 
==See Also==
 
 
{{AHSME box|year=1994|num-b=27|num-a=29}}
 
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==See Also==

Revision as of 03:49, 28 May 2021

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

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 $a=6$, so the number of solutions is $\boxed{\textbf{(C) } 2}$.

See Also

1994 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 27
Followed by
Problem 29
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