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==
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>
 
  
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 line with <math>x</math>-intercept <math>a</math> and <math>y</math>-intercept <math>b</math> is given by the equation <math>\frac{x}{a} + \frac{y}{b} = 1</math>.  We are told <math>(4,3)</math> is on the line so
  
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>.
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<cmath>\frac{4}{a} + \frac{3}{b} = 1 \implies ab - 4b - 3a = 0 \implies (a-4)(b-3)=12</cmath>
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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>.
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==See Also==
 +
 
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{{AHSME box|year=1994|num-b=27|num-a=29}}
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{{MAA Notice}}
  
 
==See Also==
 
==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
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

See Also

1994 AHSME (ProblemsAnswer KeyResources)
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

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