The number of the positive integer pairs x- y such that 1-x-1-y-1-2007 where x-y isA- 5B- 6C- 8D- 7

The correct option is D 71/x+1/y=1/2007 ⇒ (x+y)2007=xy ⇒ xy 2007x 2007y =0 (x 2007)(y 2007)=2007^2=3^4 × 223^2 The number of pairs is equal to the number of ...

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Standard XI

Accountancy

Written down Value Method(WDV)

The number of...

Question

The number of the positive integer pairs (x, y) such that $\frac{1}{x} + \frac{1}{y} = \frac{1}{2007}$ where x < y is

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Solution

The correct option is C 7
$\frac{1}{x} + \frac{1}{y} = \frac{1}{2007} \Rightarrow (x + y) 2007 = x y \Rightarrow x y - 2007 x - 2007 y = 0 (x - 2007) (y - 2007) = 2007^{2} = 3^{4} \times 223^{2}$
The number of pairs is equal to the number of divisors of $2007^{2}$ that is $(4 + 1) \times (2 + 1) = 15$
Since so required number of pairs = 7

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The number of the positive integer pairs (x, y) such that 1x+1y=12007 where x < y is

The correct option is C 71x+1y=12007⇒(x+y)2007=xy⇒xy−2007x−2007y=0(x−2007)(y−2007)=20072=34×2232 The number of pairs is equal to the number of divisors of 20072 that is (4+1)×(2+1)=15 Since so required number of pairs = 7

The number of the positive integer pairs (x, y) such that $\frac{1}{x} + \frac{1}{y} = \frac{1}{2007}$ where x < y is