Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎See Also)
m (→‎Solution)
Line 9: Line 9:
 
<cmath>\frac{4}{a} + \frac{3}{b} = 1 \implies ab - 4b - 3a = 0 \implies (a-4)(b-3)=12</cmath>
 
<cmath>\frac{4}{a} + \frac{3}{b} = 1 \implies ab - 4b - 3a = 0 \implies (a-4)(b-3)=12</cmath>
  
−
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=7</math>, so the number of solutions is <math>\boxed{\textbf{(C) } 2}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 12:24, 6 June 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 $a=7$, so the number of solutions is $\boxed{\textbf{(C) } 2}$.

See Also

1994 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 27 		Followed by
Problem 29
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1994_AHSME_Problems/Problem_28&oldid=155397"