1993 AHSME Problems/Problem 19
Problem
How many ordered pairs 
of positive integers are solutions to
![\[\frac{4}{m}+\frac{2}{n}=1?\]](http://latex.artofproblemsolving.com/b/3/b/b3b2ff9a97cace9c87b796af5a6dac9a8551413d.png)
Solution
Multiply both sides by 
to clear the denominator. Moving all the terms to the right hand side, the equation becomes 
. Adding 8 to both sides allows us to factor the equation as follows: 
. Since the problem only wants integer pairs , the pairs are given by the factors of 8, which are 1,2,4, and 8. This results in four pairs, 
and 
, which is answer
.
~42k
See also
| 1993 AHSME (Problems • Answer Key • Resources) | ||
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Followed by Problem 20 | |
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