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Number Theory Level 3

Find the number of pairs of positive integers ( x , y ) (x,y) satisfying 1 x + 1 y = 1 2013 \dfrac1x + \dfrac1y = \dfrac1{2013} .

27 36 29 42

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Mohan Nayak by mohan nayak , 19, India

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1 solution

Chew-Seong Cheong
Apr 26, 2017

1 x + 1 y = 1 2013 x + y x y = 1 2013 x y − 2013 ( x + y ) = 0 x y − 2013 ( x + y ) + 201 3 2 = 201 3 2 ( x − 2013 ) ( y − 2013 ) = 201 3 2 = 3 2 1 1 2 6 1 2 \begin{aligned} \frac 1x + \frac 1y & = \frac 1{2013} \\ \frac {x+y}{xy} & = \frac 1{2013} \\ xy - 2013(x+y) & = 0 \\ xy - 2013(x+y) + 2013^2 & = 2013^2 \\ (x-2013)(y-2013) & = 2013^2 \\ & = 3^211^261^2 \end{aligned}

The number of integer solutions of ( x , y ) (x,y) , N = ( 2 + 1 ) 3 = 27 N = (2+1)^3 = \boxed{27}

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SAME SOLUTION GREAT!!!!!!

mohan nayak - 4 years, 1 month ago

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