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Difference between revisions of "1994 AHSME Problems/Problem 28"
(→‎Solution 2)
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The only cases where <math>a</math> is:
 
The only cases where <math>a</math> is:
−
<math>\begin{cases}  
+
<cmath>\begin{cases}  
 
a-4=1 & a=5 \\
 
a-4=1 & a=5 \\
 
b-3=12 & b=15
 
b-3=12 & b=15
−
\end{cases}<cmath>
+
\end{cases}</cmath>
  
 
and
 
and
  
−
</cmath>\begin{cases}  
+
<cmath>\begin{cases}  
 
a-4=3 & a=7 \\
 
a-4=3 & a=7 \\
 
b-3=4 & b=5
 
b-3=4 & b=5
−
\end{cases}</math><math>
+
\end{cases}</cmath>
  
  
 
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
 
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
−
 
−
In the </math>xy<math>-plane, how many lines whose </math>x<math>-intercept is a positive prime number and whose </math>y<math>-intercept is a positive integer pass through the point </math>(4,3)$?
 
  
 
==See Also==
 
==See Also==

Revision as of 06:56, 28 September 2023

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 1

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

Solution 2

1984AHSMEP28.png

Let $C = (4,3)$, $DF=a$, and $AD=b$. As stated in the problem, the $x$-intercept $DF=a$ is a positive prime number, and the $y$-intercept $AD=b$ is a positive integer.

Through similar triangles, $\frac{AB}{BC}=\frac{CE}{EF}$, $\frac{b-3}{4}=\frac{3}{a-4}$, $(a-4)(b-3)=12$

The only cases where $a$ is: \[\begin{cases} a-4=1 & a=5 \\ b-3=12 & b=15 \end{cases}\]

and

\[\begin{cases} a-4=3 & a=7 \\ b-3=4 & b=5 \end{cases}\]


~isabelchen

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

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