Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: If the square root of p^2 is an integer [#permalink]

Show Tags

19 Jan 2013, 22:40

Yes, D has to be the answer.
_________________

Don't give up on yourself ever. Period. Beat it, no one wants to be defeated (My journey from 570 to 690): http://gmatclub.com/forum/beat-it-no-one-wants-to-be-defeated-journey-570-to-149968.html

Re: If the square root of p^2 is an integer, which of the follow [#permalink]

Show Tags

07 Mar 2014, 16:53

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Re: If the square root of p^2 is an integer, which of the follow [#permalink]

Show Tags

07 Mar 2014, 17:10

The first and third statements are clear. A perfect square will always have odd no. Of factors since one of the factors multiplies with itself to make that no. Also p can be a square itself so it'll give odd no. Of factors.Not necessarily even no. Of factors.

In second statement,can it be said that because 4 can be written as 2*2(even no. Of prime factors) the statement is true? Is repetition of a prime factor counted in calculating even no. Of prime factors?

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

I. p^2 has an odd number of factors II. p^2 can be expressed as the product of an even number of prime factors III. p has an even number of factors

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

The first and third statements are clear. A perfect square will always have odd no. Of factors since one of the factors multiplies with itself to make that no. Also p can be a square itself so it'll give odd no. Of factors.Not necessarily even no. Of factors.

In second statement,can it be said that because 4 can be written as 2*2(even no. Of prime factors) the statement is true? Is repetition of a prime factor counted in calculating even no. Of prime factors?

This is a flawed question. The answer to the question cannot be D.

If p=0, then none of the statements must be true.

In order for the answer to be D, the question must specify that p is a positive integer greater than 1.

In this case: The square root of p^2 is an integer --> \(\sqrt{p^2}=integer\) --> \(p=integer\).

I. p^2 has an odd number of factors --> since p is an integer, then p^2 is a perfect square. The number of factors of a positive perfect square is always odd. Thus this option must be true.

II. p^2 can be expressed as the product of an even number of prime factors. Any positive perfect square can be expressed as the product of an even number of prime factors: 4=2*2, 9=3*3, 16=2*2*2*2, 25=5*5, ... each is written as the product of even number of prime factors. Thus this option must be true.

III. p has an even number of factors --> if p itself is a perfect square, 4, 9, ... then this statement won't be true. Discard.

Re: If the square root of p^2 is an integer, which of the follow [#permalink]

Show Tags

01 Jun 2014, 23:09

Bunuel wrote:

AKG1593 wrote:

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

I. p^2 has an odd number of factors II. p^2 can be expressed as the product of an even number of prime factors III. p has an even number of factors

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

The first and third statements are clear. A perfect square will always have odd no. Of factors since one of the factors multiplies with itself to make that no. Also p can be a square itself so it'll give odd no. Of factors.Not necessarily even no. Of factors.

In second statement,can it be said that because 4 can be written as 2*2(even no. Of prime factors) the statement is true? Is repetition of a prime factor counted in calculating even no. Of prime factors?

This is a flawed question. The answer to the question cannot be D.

If p=0, then none of the statements must be true.

In order for the answer to be D, the question must specify that p is a positive integer greater than 1.

In this case: The square root of p^2 is an integer --> \(\sqrt{p^2}=integer\) --> \(p=integer\).

I. p^2 has an odd number of factors --> since p is an integer, then p^2 is a perfect square. The number of factors of a positive perfect square is always odd. Thus this option must be true.

II. p^2 can be expressed as the product of an even number of prime factors. Any positive perfect square can be expressed as the product of an even number of prime factors: 4=2*2, 9=3*3, 16=2*2*2*2, 25=5*5, ... each is written as the product of even number of prime factors. Thus this option must be true.

III. p has an even number of factors --> if p itself is a perfect square, 4, 9, ... then this statement won't be true. Discard.

Hope it helps.

Hi Bunell,

Can you please tell whether while counting the total number of factors of a perfect square, do we count 1 and the number itself?

For example :- for any number lets say 6. the total number of factors will be 1,2,3,6 .

Isn't the case with perfect squares? as if we include 1 and number itself , the total number of factors of a perfect square will be even . like for 36- 1,2,3,4,6,9,18,36

Re: If the square root of p^2 is an integer, which of the follow [#permalink]

Show Tags

01 Jun 2014, 23:52

monir6000 wrote:

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

I. p^2 has an odd number of factors II. p^2 can be expressed as the product of an even number of prime factors III. p has an even number of factors

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

I. p^2 is a perfect square and a perfect square has pairs of factors and '1' with it. The total number of factors is odd II. If p = 3 then p^2 = 9. The prime factors will always remain even as the p is an integer and the pairs have to come out of square root to make p an integer. III. This is not true always. Let us say p = 4, it has three factors: 1, 2 and 4. Hence it will not hold true always.

Hence D is the answer.
_________________

76000 Subscribers, 7 million minutes of learning delivered and 5.6 million video views

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

I. p^2 has an odd number of factors II. p^2 can be expressed as the product of an even number of prime factors III. p has an even number of factors

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

The first and third statements are clear. A perfect square will always have odd no. Of factors since one of the factors multiplies with itself to make that no. Also p can be a square itself so it'll give odd no. Of factors.Not necessarily even no. Of factors.

In second statement,can it be said that because 4 can be written as 2*2(even no. Of prime factors) the statement is true? Is repetition of a prime factor counted in calculating even no. Of prime factors?

This is a flawed question. The answer to the question cannot be D.

If p=0, then none of the statements must be true.

In order for the answer to be D, the question must specify that p is a positive integer greater than 1.

In this case: The square root of p^2 is an integer --> \(\sqrt{p^2}=integer\) --> \(p=integer\).

I. p^2 has an odd number of factors --> since p is an integer, then p^2 is a perfect square. The number of factors of a positive perfect square is always odd. Thus this option must be true.

II. p^2 can be expressed as the product of an even number of prime factors. Any positive perfect square can be expressed as the product of an even number of prime factors: 4=2*2, 9=3*3, 16=2*2*2*2, 25=5*5, ... each is written as the product of even number of prime factors. Thus this option must be true.

III. p has an even number of factors --> if p itself is a perfect square, 4, 9, ... then this statement won't be true. Discard.

Hope it helps.

Hi Bunell,

Can you please tell whether while counting the total number of factors of a perfect square, do we count 1 and the number itself?

For example :- for any number lets say 6. the total number of factors will be 1,2,3,6 .

Isn't the case with perfect squares? as if we include 1 and number itself , the total number of factors of a perfect square will be even . like for 36- 1,2,3,4,6,9,18,36

Please let me kow what i am approaching wrong

The total number of factors of any positive integer includes 1 and that integer itself. Why there should be an exception for perfect squares? Isn't a perfect square divisible by 1 and itself?

The factors of 36 are: 1, 2, 3, 4, 6, 9, 12, 18 and 36 --> 9 factors.
_________________

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

I. p^2 has an odd number of factors II. p^2 can be expressed as the product of an even number of prime factors III. p has an even number of factors

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

I. p^2 is a perfect square and a perfect square has pairs of factors and '1' with it. The total number of factors is odd II. If p = 3 then p^2 = 9. The prime factors will always remain even as the p is an integer and the pairs have to come out of square root to make p an integer. III. This is not true always. Let us say p = 4, it has three factors: 1, 2 and 4. Hence it will not hold true always.

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Happy 2017! Here is another update, 7 months later. With this pace I might add only one more post before the end of the GSB! However, I promised that...