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Note: a perfect square, is an integer that can be written as the square of some other integer. For example \(16=4^2\), is a perfect square. Similarly, a perfect cube is an integer that can be written as the cube of some other integer. For example, \(27=3^3\) is a perfect cube.

Make prime factorization of 2,700: \(xyz=2,700=2^2*3^3*5^2\).

(1) \(y\) is an even perfect square and \(z\) is an odd perfect cube. If \(y\) is either \(2^2\) or \(2^2*5^2\) and \(z=3^3= \text{odd perfect square}\) then \(x\) must be a perfect square which makes \(\sqrt{x}\) an integer: \(x=5^2\) or \(x=1\). But if \(z=1^3= \text{odd perfect cube}\) then \(x\) could be \(3^3\) which makes \(\sqrt{x}\) not an integer. Not sufficient.

(2) \(\sqrt{z}\) is not an integer. Clearly insufficient.

(1)+(2) As from (1) \(\sqrt{z} \ne integer\) then \(z \ne 1\), therefore it must be \(3^3\) (from 1), so \(x\) must be a perfect square which makes \(\sqrt{x}\) an integer: \(x=5^2\) or \(x=1\). Sufficient.

Note: a perfect square, is an integer that can be written as the square of some other integer. For example \(16=4^2\), is a perfect square. Similarly, a perfect cube is an integer that can be written as the cube of some other integer. For example, \(27=3^3\) is a perfect cube.

Make prime factorization of 2,700: \(xyz=2,700=2^2*3^3*5^2\).

(1) \(y\) is an even perfect square and \(z\) is an odd perfect cube. If \(y\) is either \(2^2\) or \(2^2*5^2\) and \(z=3^3= \text{odd perfect square}\) then \(x\) must be a perfect square which makes \(\sqrt{x}\) an integer: \(x=5^2\) or \(x=1\). But if \(z=1^3= \text{odd perfect cube}\) then \(x\) could be \(3^3\) which makes \(\sqrt{x}\) not an integer. Not sufficient.

(2) \(\sqrt{z}\) is not an integer. Clearly insufficient.

(1)+(2) As from (1) \(\sqrt{z} \ne integer\) then \(z \ne 1\), therefore it must be \(3^3\) (from 1), so \(x\) must be a perfect square which makes \(\sqrt{x}\) an integer: \(x=5^2\) or \(x=1\). Sufficient.

Answer: C

can you please explain 2) \sqrt{z} is not an integer. Clearly insufficient. and sqrt{z} ne integer then z ne 1

Note: a perfect square, is an integer that can be written as the square of some other integer. For example \(16=4^2\), is a perfect square. Similarly, a perfect cube is an integer that can be written as the cube of some other integer. For example, \(27=3^3\) is a perfect cube.

Make prime factorization of 2,700: \(xyz=2,700=2^2*3^3*5^2\).

(1) \(y\) is an even perfect square and \(z\) is an odd perfect cube. If \(y\) is either \(2^2\) or \(2^2*5^2\) and \(z=3^3= \text{odd perfect square}\) then \(x\) must be a perfect square which makes \(\sqrt{x}\) an integer: \(x=5^2\) or \(x=1\). But if \(z=1^3= \text{odd perfect cube}\) then \(x\) could be \(3^3\) which makes \(\sqrt{x}\) not an integer. Not sufficient.

(2) \(\sqrt{z}\) is not an integer. Clearly insufficient.

(1)+(2) As from (1) \(\sqrt{z} \ne integer\) then \(z \ne 1\), therefore it must be \(3^3\) (from 1), so \(x\) must be a perfect square which makes \(\sqrt{x}\) an integer: \(x=5^2\) or \(x=1\). Sufficient.

Answer: C

can you please explain 2) \sqrt{z} is not an integer. Clearly insufficient. and sqrt{z} ne integer then z ne 1

\(z \ne 1\) because if it were 1, then \(\sqrt{z}\) would be an integer and that would violate the second statement.
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Why is the answer not D) Why is 2) not sufficient, if the square root of Z is not an integer that means that Z must be 3^ 3 and x must be either 5^2 or 2^2 and the square root of either of those number is an integer.

Make prime factorization of 2,700: xyz=2,700=22∗33∗52xyz=2,700=2^2*3^3*5^2.

Why is the answer not D) Why is 2) not sufficient, if the square root of Z is not an integer that means that Z must be 3^ 3 and x must be either 5^2 or 2^2 and the square root of either of those number is an integer.

Make prime factorization of 2,700: xyz=2,700=22∗33∗52xyz=2,700=2^2*3^3*5^2.

Why is the answer not D) Why is 2) not sufficient, if the square root of Z is not an integer that means that Z must be 3^ 3 and x must be either 5^2 or 2^2 and the square root of either of those number is an integer.

Make prime factorization of 2,700: xyz=2,700=22∗33∗52xyz=2,700=2^2*3^3*5^2.

would like as well answer to this question.

There are other cases possible. For example, z=3, y=2*3*5^2 and x=2.
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