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A)$5$ B)$6$ C)$7$ D) $8$

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Verified

We are given that,

$ \Rightarrow $ $\dfrac{1}{{\text{x}}} + \dfrac{1}{{\text{y}}} = \dfrac{1}{{2007}}$

We have to find the number of positive integers (x, y) such that ${\text{x < y}}$

So first solve the given fractions. On taking LCM, we get

$ \Rightarrow \dfrac{{{\text{x + y}}}}{{{\text{xy}}}} = \dfrac{1}{{2007}}$

On cross multiplication, we get

$ \Rightarrow {\text{xy = 2007}}\left( {{\text{x}} + {\text{y}}} \right) \Rightarrow {\text{xy}} - {\text{2007}}\left( {{\text{x}} + {\text{y}}} \right) = 0$

Now, to find the values of x and y , we can add the ${2007^2}$ on both sides so we can make factors of x and y.

$ \Rightarrow {\text{xy}} - {\text{2007}}\left( {{\text{x}} + {\text{y}}} \right) + {2007^2} = {2007^2}$

On simplifying we get,

$

\Rightarrow {\text{xy}} - 2007{\text{x}} - 2007{\text{y}} + {2007^2} = {2007^2} \\

\Rightarrow {\text{x}}\left( {{\text{y}} - 2007} \right) - 2007\left( {{\text{y}} - 2007} \right) = {2007^2} \\

\Rightarrow \left( {{\text{x}} - 2007} \right)\left( {{\text{y - }}2007} \right) = {2007^2} \\

$

Now we can take A=${\text{x}} - 2007$ and B=${\text{y}} - 2007$ , then the equation becomes

$ \Rightarrow {\text{AB = }}{2007^2}$

Now from the equation it is clear that the number of solutions of this equation will be equal to the number of factors of ${2007^2}$.So we can write $2007 = 9 \times 223$ $ = {3^2} \times 223$ .It canāt be further simplified as $3$ and $223$ are both prime numbers.

Then we can write ${2007^2}$$ = {3^4} \times {223^2}$

Here, we are finding the number of factors in the form of prime factors hence,

$ \Rightarrow {\text{number of factor = }}\left( {4 + 1} \right)\left( {2 + 1} \right) = 5 \times 3 = 15$

This is the total number of factors of ${2007^2}$.Now in all the $15$ cases there will be only one case when A=B=$2007$ as $\left\{ {{\text{A}} \times {\text{B = 200}}{{\text{7}}^2}} \right\}$ .So in other 14 case either A>B or B>A so there will be 7 cases where $A>B$ which means $x>y$.

The number of divisor or factors of a number ānā in the form of prime factors \[ \Rightarrow {\text{a}}_1^{{{\text{P}}_1}}.{\text{a}}_2^{{{\text{P}}_2}}...{\text{a}}_{\text{n}}^{{\text{Pn}}} = \left( {{{\text{P}}_1} + 1} \right)\left( {{{\text{P}}_2} + 1} \right)...\left( {{{\text{P}}_{\text{n}}} + 1} \right)\]