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If the square root of p^2 is an integer, which of the follow

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If the square root of p^2 is an integer, which of the follow [#permalink]  19 Jan 2013, 20:31
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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
[Reveal] Spoiler: OA

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Last edited by monir6000 on 20 Jan 2013, 19:41, edited 2 times in total.
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Re: If the square root of p^2 is an integer [#permalink]  19 Jan 2013, 20:48
Answer should be D and not B.
Every square has odd number of factors.
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Re: If the square root of p^2 is an integer [#permalink]  19 Jan 2013, 22:40
Yes, D has to be the answer.
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Re: If the square root of p^2 is an integer, which of the follow [#permalink]  07 Mar 2014, 16:53
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Re: If the square root of p^2 is an integer, which of the follow [#permalink]  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?

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Re: If the square root of p^2 is an integer, which of the follow [#permalink]  08 Mar 2014, 05:29
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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.
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Re: If the square root of p^2 is an integer, which of the follow [#permalink]  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

Please let me kow what i am approaching wrong
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Re: If the square root of p^2 is an integer, which of the follow [#permalink]  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.

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Re: If the square root of p^2 is an integer, which of the follow [#permalink]  02 Jun 2014, 00:04
Expert's post
Manik12345 wrote:
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

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.
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Re: If the square root of p^2 is an integer, which of the follow [#permalink]  02 Jun 2014, 00:06
Expert's post
PerfectScores wrote:
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.

This would be correct if we were told that p is a positive integer greater than 1. Check here: if-the-square-root-of-p-2-is-an-integer-which-of-the-follow-146066.html#p1341427
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Re: If the square root of p^2 is an integer, which of the follow   [#permalink] 02 Jun 2014, 00:06
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