Difference between revisions of "1993 AHSME Problems/Problem 19"
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Multiply both sides by <math>mn</math> to clear the denominator. Moving all the terms to the right hand side, the equation becomes <math>0 = mn-4n-2m</math>. Adding 8 to both sides allows us to factor the equation as follows: <math>(m-4)(n-2) = 8</math>. Since the problem only wants integer pairs <math>(m,n)</math>, the pairs are given by the factors of 8, which are 1,2,4, and 8. This results in four pairs, <math>(5,10), (6,6), (8,4),</math> and <math>(12,3)</math>, which is answer | Multiply both sides by <math>mn</math> to clear the denominator. Moving all the terms to the right hand side, the equation becomes <math>0 = mn-4n-2m</math>. Adding 8 to both sides allows us to factor the equation as follows: <math>(m-4)(n-2) = 8</math>. Since the problem only wants integer pairs <math>(m,n)</math>, the pairs are given by the factors of 8, which are 1,2,4, and 8. This results in four pairs, <math>(5,10), (6,6), (8,4),</math> and <math>(12,3)</math>, which is answer | ||
<math>\fbox{D}</math>. | <math>\fbox{D}</math>. | ||
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| + | ~42k | ||
== See also == | == See also == | ||
Revision as of 22:30, 22 December 2020
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) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
