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Question

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

A
5
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B
6
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C
7
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D
8
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Solution

The correct option is B 7

x+yxy=12007

xy2007(x+y)=0

Adding 20072 to both sides, we get

xy2007(x+y)+20072=20072

(x2007)(y2007)=20072

Let x2007=A and y2007=B

The equation becomes AB=20072

Number of solutions of above equation is equal to number of factors of 20072

20072=34×2232

Hence, number of factors of 20072 is (4+1)(2+1)=15

In one case A=B=2007

Of the remaining 14 cases, half of the case A>B and remaining half A<B

Accordingly we get 7 cases, where x<y.


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