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Math Expert V
Joined: 02 Sep 2009
Posts: 53796
If the square root of p^2 is an integer greater than 1, which of the  [#permalink]

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Question Stats: 55% (01:59) correct 45% (02:17) wrong based on 220 sessions

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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

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Posts: 7434
Re: If the square root of p^2 is an integer greater than 1, which of the  [#permalink]

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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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1
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.

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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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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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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: If the square root of p^2 is an integer greater than 1, which of the  [#permalink]

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Hi All,

This Roman Numeral question is based on a series of Number Property rules, but you don't need to know the rules to get the correct answer - you can TEST VALUES and do some brute-force math.

From the prompt, we know that P is a positive INTEGER greater than 1. We're asked which of the 3 Roman Numerals MUST be true. In most Roman Numeral questions, the 'key' is to DISPROVE the Roman Numerals so that we can quickly eliminate answer choices. Here though, we're going to prove that patterns exist.

Since P is a positive integer, we know that P^2 is a perfect square.

I. P^2 has an odd number of positive factors

IF...
P = 2, P^2 = 4 and the factors are 1, 2 and 4... so there IS an odd number of factors
P = 3, P^2 = 9 and the factors are 1, 3 and 9... so there IS an odd number of factors
P = 4, P^2 = 16 and the factors are 1, 2, 4, 8 and 16... so there IS an odd number of factors
Notice the pattern here. Since P^2 is a perfect square, there will ALWAYS be an odd number of factors, so Roman Numeral 1 IS true.
Eliminate Answers B, C and E.

From the answers that remain, we only have to deal with Roman Numeral II.

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

Using the same examples from Roman Numeral I, you can prove this pattern too:
P = 2, P^2 = 4 and we can get to 4 by multiplying (2)(2)... an even number of positive prime factors
P = 3, P^2 = 9 and we can get to 9 by multiplying (3)(3)... an even number of positive prime factors
P = 4, P^2 = 16 and we can get to 16 by multiplying (2)(2(2)(2))... an even number of positive prime factors
Since P^2 is a perfect square there will ALWAYS be a product of positive prime factors that will end in P^2, so Roman Numeral 2 IS true.

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*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***** Re: If the square root of p^2 is an integer greater than 1, which of the   [#permalink] 10 Jan 2018, 19:45
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