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How many different pairs $(a,b)$ of positive integers are there such that $a\leq b$ and $1/a+1/b=1/9$

  1. None of these
  2. $2$
  3. $0$
  4. $1$
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Given that, $\dfrac{1}{a}+\dfrac{1}{b} = \dfrac{1}{9}\;; \;a,b\in \mathbb{Z}^{+} $ and $a\leq b$

$\Rightarrow \dfrac{a+b}{ab} = \dfrac{1}{9}$

$\Rightarrow 9a+9b = ab$

$\Rightarrow 9a+9b-ab = 0$

$\Rightarrow ab-9a-9b = 0$

$\Rightarrow ab-9a-9b+81 = 81$

$\Rightarrow (a-9)(b-9) = 81$     

We can factorize $81$ such that $a-9\leq b-9 \Rightarrow \boxed{a\leq b}$     

$ \qquad \qquad \begin{array} {ccc} \underline{a-9}& \leq  & \underline{b-9} \\  1 & & 81 \\ 3 & & 27 \\ 9 &  & 9\end{array}$

$\therefore$ Only three pairs are possible.

Correct Answer $:\text{A}$
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