If z is a positive integer is. . . .

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If z is a positive integer is. . . .

by VJesus12 » Sun Oct 15, 2017 5:58 pm
$$If\ z\ is\ a\ positive\ integer,\ is\ \sqrt{z}\ an\ integer?$$
$$(1)\ \sqrt{xz}\ is\ an\ integer$$
$$(2)\ x=z^3$$

The OA is E.

I think A is the correct answer.

If $$\sqrt{xz},$$ then $$\sqrt{x}\sqrt{z}$$ is an integer, which implies that $$\sqrt{x} \ and\ \ \sqrt{z}$$ are integers.

Am I wrong?
Source: — Data Sufficiency |

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by Jay@ManhattanReview » Mon Oct 16, 2017 10:32 pm
VJesus12 wrote:$$If\ z\ is\ a\ positive\ integer,\ is\ \sqrt{z}\ an\ integer?$$
$$(1)\ \sqrt{xz}\ is\ an\ integer$$
$$(2)\ x=z^3$$

The OA is E.

I think A is the correct answer.

If $$\sqrt{xz},$$ then $$\sqrt{x}\sqrt{z}$$ is an integer, which implies that $$\sqrt{x} \ and\ \ \sqrt{z}$$ are integers.

Am I wrong?
Hi VJesus12,

Statement 1 is not sufficient.

If √(xz) is an integer then √x.√z is an integer, which DOES NOT imply that √x and √z are integers.

See how...

Say x=2 & z =2, then √x.√z = √2.√2 = 2 is an integer, but √z = √2 is not an integer.

Hope this helps!

-Jay

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