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# Find the number of positive integer pairs(x, y) such that $\dfrac{1}{{\text{x}}} + \dfrac{1}{{\text{y}}} = \dfrac{1}{{2007}},{\text{x < y}}$ -A)$5$ B)$6$ C)$7$ D) $8$

Answer Verified
Hint: Take LCM of the given fractions and solve to form an equation. Add the square of $2007$ in the equation on both sides and simplify. Then find the factors of ${2007^2}$ because the number of solutions of the equation will be equal to the number of factors.

Complete step-by-step answer:
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 can’t 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 $x>y$.

So, the number of positive integers of pairs (x, y) such that ${\text{x < y}}$ will be 7.

Note: Here, we have used the formula for the number of factors in the form of prime factors which is-
The number of divisor or factors of a number “n” in the form of prime factors $\Rightarrow {\text{a}}_1^{{{\text{P}}_1}}.{\text{a}}_2^{{{\text{P}}_2}}...{\text{a}}_{\text{n}}^{{\text{Pn}}} = \left( {{{\text{P}}_1} + 1} \right)\left( {{{\text{P}}_2} + 1} \right)...\left( {{{\text{P}}_{\text{n}}} + 1} \right)$