# 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> | ||

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+ | 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 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>. |

## Revision as of 04:36, 18 February 2018

## 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

Let the line be , with being a positive integer. Then we have so . Now the -intercept is

We need the -intercept to be positive, so and together imply , so if the -intercept is to be an integer, must be a positive factor of . Hence we can get , , , , , and , and the only ones of these that are prime are and .

The first case gives and the second case gives , and both of these satisfy all the conditions of the problem, so the number of solutions is .