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If the square root of p^2 is an integer greater than 1, which of the

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If the square root of p^2 is an integer greater than 1, which of the [#permalink]

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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
[Reveal] Spoiler: OA

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Re: If the square root of p^2 is an integer greater than 1, which of the [#permalink]

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New post 04 Apr 2016, 08:16
Bunuel wrote:
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


It is given that \(\sqrt{p^2}\)>1 and is an integer..
Also \(\sqrt{p^2}\)=p
so p^2 is a perfect square and p is an integer>1..

lets see three statements:-
I. p^2 has an odd number of positive factors-- TRUE
A perfect square has an Odd number of perfect square ..
WHY? Because all factors will have a different integer as a pair except p..
p*p=p^2.. so that will be just one factor


II. p^2 can be expressed as the product of an even number of positive prime factors
again TRUE
25 =5*5 ; 49= 7*7 ; 36 = 2*2*3*3


III. p has an even number of positive factors..
Since it uses 'HAS', if we find one which does not follow the statement is not true..
we already know that A perfect square has an Odd number of perfect square ..
so if p is a perfetct square itself ans is No


ONLY I and II are true
D
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Re: If the square root of p^2 is an integer greater than 1, which of the [#permalink]

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New post 04 Apr 2016, 14:09
Bunuel wrote:
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


We are told that \(\sqrt{p^2}>1\), and it is an integer
\(\sqrt{p^2}>1\) is the definition of the absolute value of p, so the question is saying that \(|p|>1\), and is an integer.

The question also asks which of the following MUST be true, so if there is even one case where the statement is not true, then that statement is out.

Let's take a look at each statement:

I. \(p^2\) has an odd number of positive factors

Since \(p\) is an integer, \(p^2\) is a perfect square. Is it possible for a perfect square to have an even number of positive factors? No, it isn't. All perfect squares have an odd number of positive factors. Why? Factors always come in pairs, but for a perfect square, \(p^2\), the factor \(p\) is paired with itself, and so is only counted once.

Example: if \(p^2=36\), the factor pairs are
1*36
2*18
3*12
4*9
6*6 <-- 6 is paired with itself, and so it is only counted as one factor
36 has 9 factors (1, 2, 3, 4, 6, 9, 12, 18, 36). This is true of every perfect square. Statement I MUST be true.


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

This also must be true. The prime factors of a number \(n\) can be expressed in the form \(p_1^{a_1}, p_2^{a_2}, p_3^{a_3}\), etc, where \(p_1, p_2, p_3\) are the prime factors of \(n\), and \(a_1, a_2, a_3\) are the number of times those factors appear. For example, if \(p^2 = 36\), then the prime factors of \(p^2\) are 2^2, 3^2. So 36 can be expressed as the product of an even number of positive prime factors: 36 = 2*2*3*3 --> the product of four prime factors.

Every perfect square can be expressed as the product of an even number of positive prime factors because the powers on each prime factor (\(a_1, a_2\), etc.) will always be even. We know this because when a number is squared, all the prime factors are squared, i.e. the exponent on each prime factor is doubled. for example, if a number \(n\) with prime factors \(3^3, 47^7, 151^{127}\) is squared, then the prime factors of \(n^2\) will be \(3^6, 47^{14}, 151^{254}\) --> all the exponents have been doubled, thus \(n^2\) can always be expressed as the product of an even number of prime factors.


III. p has an even number of positive factors

We are given that \(\sqrt{p^2}\) is an integer >1, but how does that restrict the value of \(p\)? \(\sqrt{p^2}\) is equivalent to saying \(|p|\) is an integer >1. So \(p\) could be any integer \(\leq\)-2, or \(\geq\)2. Are there any integers in that range that have an odd number of positive factors? Based on our analysis of statement I, obviously yes. If \(p\) is itself a perfect square, then it will have an odd number of positive factors, and statement III would not be true.

Thus statements I and II must be true and statement III is not necessarily true.

Answer: D
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Re: If the square root of p^2 is an integer greater than 1, which of the [#permalink]

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New post 18 Aug 2017, 02:10
Number of factors of a given number cab be calculated as....
If N = \(a^{l}\)*\(b^{q}\)*\(c^{r}\).....
where a,b,c... are prime numbers and p,q,r,....are their powers, then number of factors of the number N can be calculated as
(l+1)*(q+1)*(r+1).....
For example: Let N = 24 = \(2^{3}\)*\(3^{1}\)
Here, a= 2, b=3, l=3, q=1
Thus number of factors of N = (l+1)*(q+1) = (3+1)*(1+1) = 4*2 = 8


Using same concept: If \(\sqrt{P^{2}}\) = P
then, P^{2} will have number of factors as (2+1) = 3, odd. Statement I is correct
Thus we can eliminate options B,C & E
Now statement II: any number P can be written as products of its prime factors P= \(a^{l}\)*\(b^{q}\)*\(c^{r}\)..... (where a,b,c ... are prime numbers)
Squaring p means squaring powers of all prime numbers i.e. \(P^{2}\)= (\(a^{l}\)*\(b^{q}\)*\(c^{r}\).....)^{2}
Thus all the powers are getting multiplied by 2 and thus they become even.Statement II is correct
Thus we can eliminate option A also

Thus answer is option D
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Re: If the square root of p^2 is an integer greater than 1, which of the   [#permalink] 18 Aug 2017, 02:10
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