Difference between revisions of "1992 AHSME Problems/Problem 13"
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== Solution == | == Solution == | ||
| − | <math>\fbox{C}</math> | + | <math>\fbox{C}</math> We can rewrite the left-hand side as <math>\frac{a^2b+a}{b+ab^2}</math> by multiplying the numerator and denominator by <math>ab</math>. Now we can multiply up by the denominator to give <math>a^2b+a = 13b + 13ab^2 \implies ab(a-13b)=13b-a = -(a-13b) \implies (ab+1)(a-13b) = 0</math>. Now, as <math>a</math> and <math>b</math> are positive, we cannot have <math>ab+1=0</math>, so we must have <math>a-13b=0 \implies a=13b</math>, so the solutions are <math>(13,1), (26,2), (39,3), (52,4), (65,5), (78,6), (91,7)</math>, which is 7 ordered pairs. |
== See also == | == See also == | ||
Revision as of 01:40, 20 February 2018
Problem
How many pairs of positive integers (a,b) with 
satisfy the equation
![\[\frac{a+b^{-1}}{a^{-1}+b}=13?\]](http://latex.artofproblemsolving.com/e/1/4/e141f8438cc8bcf014c0f636e2f4433b1da2d5d3.png)
Solution
We can rewrite the left-hand side as 
by multiplying the numerator and denominator by 
. Now we can multiply up by the denominator to give 
. Now, as 
and 
are positive, we cannot have 
, so we must have 
, so the solutions are 
, which is 7 ordered pairs.
See also
| 1992 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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. 
