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# Number Theory

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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