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 1== | ||
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 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 | ||
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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=7</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>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | [[File:1984AHSMEP28.png|500px|center]] | ||
+ | |||
+ | Let <math>C = (4,3)</math>, <math>DF=a</math>, and <math>AD=b</math>. As stated in the problem, the <math>x</math>-intercept <math>DF=a</math> is a positive prime number, and the <math>y</math>-intercept <math>AD=b</math> is a positive integer. | ||
+ | |||
+ | Through similar triangles, <math>\frac{AB}{BC}=\frac{CE}{EF}</math>, <math>\frac{b-3}{4}=\frac{3}{a-4}</math>, <math>(a-4)(b-3)=12</math> | ||
+ | |||
+ | The only cases where <math>a</math> is: | ||
+ | <math>\begin{cases} | ||
+ | a-4=1 & a=5 \\ | ||
+ | b-3=12 & b=15 | ||
+ | \end{cases}<cmath> | ||
+ | |||
+ | and | ||
+ | |||
+ | </cmath>\begin{cases} | ||
+ | a-4=3 & a=7 \\ | ||
+ | b-3=4 & b=5 | ||
+ | \end{cases}</math><math> | ||
+ | |||
+ | |||
+ | ~[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 06:55, 28 September 2023
Contents
Problem
In the -plane, how many lines whose -intercept is a positive prime number and whose -intercept is a positive integer pass through the point ?
Solution 1
The line with -intercept and -intercept is given by the equation . We are told is on the line so
Since and are integers, this equation holds only if is a factor of . The factors are which means must be one of . The only members of this list which are prime are and , so the number of solutions is .
Solution 2
Let , , and . As stated in the problem, the -intercept is a positive prime number, and the -intercept is a positive integer.
Through similar triangles, , ,
The only cases where is: $\begin{cases} a-4=1 & a=5 \\ b-3=12 & b=15 \end{cases}<cmath>
and
</cmath>\begin{cases} a-4=3 & a=7 \\ b-3=4 & b=5 \end{cases}$ (Error compiling LaTeX. Unknown error_msg)$~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
In the$ (Error compiling LaTeX. Unknown error_msg)xyxy(4,3)$?
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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