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

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\text{(E) } 13</math>
 
\text{(E) } 13</math>
  
== Solution ==
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== Solution 1 ==
 
<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.
 
<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.
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== Solution 2 ==
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The equation is equivalent to <math>\frac{a+\frac{1}{b}}{\frac{1}{a}+b}=13</math> or <math>\frac{\frac{ab+1}{b}}{\frac{ab+1}{a}}=13</math>. The <math>ab+1</math>s cancel out leaving us with <math>\frac{a}{b}=13</math>. This is asking us for the positive multiples of <math>13</math> up to <math>100</math>, and we know that <math>13 \cdot 7 = 91</math>, so our answer is <math>\fbox{C}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 21:44, 4 April 2021

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 1

$\fbox{C}$ 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.

Solution 2

The equation is equivalent to $\frac{a+\frac{1}{b}}{\frac{1}{a}+b}=13$ or $\frac{\frac{ab+1}{b}}{\frac{ab+1}{a}}=13$. The $ab+1$s cancel out leaving us with $\frac{a}{b}=13$. This is asking us for the positive multiples of $13$ up to $100$, and we know that $13 \cdot 7 = 91$, so our answer is $\fbox{C}$.

See also

1992 AHSME (ProblemsAnswer KeyResources)
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
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