The prime factorization of 2007 is 3^2*223. How many ordered pairs of positive integers (x,y) satisfy the equation xy^2=2007?
+678 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
