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If is an integer and xy2 = 36, how many values are possible [#permalink]
Since \(\sqrt{x}\) = integer, the value of \(x\) should be a perfect square. At the same time, the value of \(y\) should be such that its square when multiplied by \(x\) gives \(36\)

The possible combinations are

\(x \times y^2 = 36\)

\(1 \times 36 => x=1, \sqrt{x}=1, y^2=36, y=6 \text{ or } -6\)

\(4 \times 9 => x=4, \sqrt{x}=2, y^2=9, y=3 \text{ or } -3\)

\(9 \times 4 => x=9, \sqrt{x}=3, y^2=4, y=2 \text{ or } -2\)

\(36 \times 1 => x=36, \sqrt{x}=6, y^2=1, y=1 \text{ or } -1\)

Therefore the possible values of \(y\) are \(1,-1,2,-2,3,-3,6,-6\).

There are Eight possible values of \(y\).

The answer is E.
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