1994 AHSME Problems/Problem 28: Difference between revisions
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<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= | 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
-plane, how many lines whose
-intercept is a positive prime number and whose
-intercept is a positive integer pass through the point
?
Solution
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
.
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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