If the square root of $p^2$ is an integer greater than 1, which of the following must be true?

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Vincen: Legendary Member; Posts: 2898; Joined: Thu Sep 07, 2017 2:49 pmFollowed by: 5 members; Thanked: 6 times

by Vincen » Mon May 18, 2020 7:03 am

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If the square root of $p^2$ is an integer greater than 1, which of the following must be true?

I. $p^2$ has an odd number of positive factors

II. $p^2$ can be expressed as the product of an even number of positive prime factors

III. $p$ has an even number of positive factors

A. I
B. II
C. III
D. I and II
E. II and III

[spoiler]OA=D[/spoiler]

Source: Manhattan GMAT

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Source: Beat The GMAT — Problem Solving | Mon May 18, 2020 7:03 am

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8152; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

Re: If the square root of $p^2$ is an integer greater than 1, which of the following must be true?

by Scott@TargetTestPrep » Sun May 24, 2020 4:57 am

Vincen wrote: ↑
Mon May 18, 2020 7:03 am
If the square root of $p^2$ is an integer greater than 1, which of the following must be true?

I. $p^2$ has an odd number of positive factors

II. $p^2$ can be expressed as the product of an even number of positive prime factors

III. $p$ has an even number of positive factors

A. I
B. II
C. III
D. I and II
E. II and III

[spoiler]OA=D[/spoiler]

Since p^2 > 1, p > 1 or p < -1. Let’s consider p > 1 first.

If p > 1, then p^2 always has an odd number of positive factors. So statement I is true. Statement II is also true since every distinct prime factor of p^2 must come in pairs or an even amount. However, statement III is not necessarily true since p itself can be a perfect square so p could have an odd number of positive factors.

Now, let’s consider p < -1.

If p < -1, statements I and II will still be true since p^2 = |p|^2 (notice that |p| > 1, so it behaves the same as p > 1). We don’t have to consider statement III because it’s proven false when p > 1 already.

Answer: D

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Founder and CEO
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