Question

Find the number of ordered pairs of positive integers (x,ythat satisfy the equation 1/x+1/y =1/2004
​

Answer 1

Answer:

(x-2004)(y-2004)=2004²=2⁴ 3² =167²

Answer 2

1/x+1/y =1/2004

​⇒$\frac{x+y}{xy}$ = $\frac{1}{2004}$

⇒ 2004 ( x+y ) = xy

⇒ 2004x - xy + 2004 y =0

⇒2004x - xy + 2004y - (2004)² = - (m 2004)²

⇒ x ( 2004 -y) + 2004 ( y-2004) = - ( 2004)²

⇒ - x ( y - 2004)+ 2004 ( y - 2004)= - ( 2004)²

⇒( y-2004) ( -x + 2004) = - ( 2004)²

⇒ ( x- 2004) ( y-2004) = ( 2004)²

( 2004)² = 2⁴ + 3² + 167²

             = ( 4 +1) ( 2+1) ( 2+1)

             = (5× 3×3) = 45

There are exactly ( 5× 3×3 )= 45 positive integer divisors of this number.

Ans :- 45.

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