Difference between revisions of "1992 AHSME Problems/Problem 13"

(Created page with "== Problem == How many pairs of positive integers (a,b) with <math>a+b\le 100</math> satisfy the equation <cmath>\frac{a+b^{-1}}{a^{-1}+b}=13?</cmath> <math>\text{(A) } 1\quad...")
 
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== Solution ==
 
== Solution ==
<math>\fbox{C}</math>
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<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 02:40, 20 February 2018

Problem

How many pairs of positive integers (a,b) with $a+b\le 100$ satisfy the equation

\[\frac{a+b^{-1}}{a^{-1}+b}=13?\]

$\text{(A) } 1\quad \text{(B) } 5\quad \text{(C) } 7\quad \text{(D) } 9\quad \text{(E) } 13$

Solution

$\fbox{C}$ We can rewrite the left-hand side as $\frac{a^2b+a}{b+ab^2}$ by multiplying the numerator and denominator by $ab$. Now we can multiply up by the denominator to give $a^2b+a = 13b + 13ab^2 \implies ab(a-13b)=13b-a = -(a-13b) \implies (ab+1)(a-13b) = 0$. Now, as $a$ and $b$ are positive, we cannot have $ab+1=0$, so we must have $a-13b=0 \implies a=13b$, so the solutions are $(13,1), (26,2), (39,3), (52,4), (65,5), (78,6), (91,7)$, which is 7 ordered pairs.

See also

1992 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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