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The prime factorization of 2007 is 3^2*223. How many ordered pairs of positive integers (x,y) satisfy the equation xy^2=2007?

 Nov 6, 2021
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Hello Guest,

 

The prime factorization of 2007 is 3^2*223. How many ordered pairs of positive integers (x,y) satisfy the equation xy^2=2007?

 

\(xy^2=2007\)

 

\(x = 223\)

\(y = \pm \mbox { } 3\)

 

\(x = 2007\)

\(y= \pm \mbox { } 1\)

 

These are the onliest possibilities, \((x, \mbox { } y = 223, \mbox { } \pm \mbox { } 3)\) and \((x, \mbox { } y = 2007, \mbox { } \pm \mbox { } 1)\) .

 

Straight

 Nov 10, 2021

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