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

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

by rsarashi » Sun Jul 16, 2017 12:03 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

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by DavidG@VeritasPrep » Sun Jul 16, 2017 4:43 am
rsarashi 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

OAD[spoiler][/spoiler]
You could always pick some easy numbers. Say p = 4; 4^2 = 16
I) True. In this case there are five factors of p^2, or 16: 1, 2, 4, 8, 16
II) True. 16 is 2 * 2 * 2 * 2, so it can be expressed as an even number of prime factors.
III) False. 4 has three factors: 1, 2, 4

So I and II work in this instance.

Say p = 6.
I) True. There are nine factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36
II) True: 36 is 2*2*3*3
III)No need to test again, we already know that it need not be true.

No matter what we pick, I and II will be true. The answer is D. (Note that we don't have to bother testing negative numbers here as the true statements all involve p^2.)
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by DavidG@VeritasPrep » Sun Jul 16, 2017 4:49 am
Or we can use some logic/number properties.

- A perfect square will always have an odd number of factors. (we can easily find the number of factors of a number by first finding the prime factorization, adding '1' to each exponent, and then multiplying the results together. More abstractly, if the prime factorization of a number is (x^a) * (y^b) *(z^c) where x, y, and z are all prime bases, the number of factors will be (a +1)(b+1)(c+1). In a perfect square, a, b, and c must all be even, so when we add one, all the numbers will be odd.) Thus statement 1 must be true.

- A perfect square must have an EVEN number of prime bases. (Note: not unique bases. Rather the exponent of each prime base must be an even number.) If we start with p, and the square it, we're effectively multiplying the number of prime bases by 2, and any integer multiplied by 2 is EVEN. Thus statement II must be true.
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by DavidG@VeritasPrep » Sun Jul 16, 2017 4:50 am
One last point: as soon as you confirm that statements I and II are true, there's no reason to bother testing III, as there's no option among the answer choices that claims all the statements are true. Always keep one eye on the answer choices as you work to save yourself from doing unnecessary analysis!
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by [email protected] » Sun Jul 16, 2017 4:57 am
Hi rsarashi,

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.
Eliminate Answer A.

Final Answer: D

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by Ian Stewart » Mon Jul 17, 2017 7:14 am
You don't need to know any particular number theory methods to see why a perfect square always has an odd number of divisors. If you look at a number that is not a square first, all of its divisors will be in pairs. Take the number 12, say:

1 * 12
2 * 6
3 * 4

We have six divisors, which I've listed in pairs that give us 12 as a product. But if we take a perfect square, one divisor won't be in a pair. Take 16, say:

1 * 16
2 * 8
4^2

We have an odd number of divisors, because we multiply 4 by itself to get 16, so the '4' isn't paired with a different divisor in the list above. You'll only ever have an unpaired divisor when you list the divisors of a perfect square, so only perfect squares can have an odd number of divisors, and all other numbers have an even number of divisors.
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by Jay@ManhattanReview » Mon Jul 17, 2017 9:42 pm
rsarashi 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

OAD[spoiler][/spoiler]
I think there is something missing in the question.

We have √p^2 > 1

=> p^2: {2, 3, 4, 5, ...}

=> |p| > 1

=> p < -1 OR p > 1

=> p is not necessarily an integer at all. If it is so, then none of the statements is true.

For option D to be the correct answer, the question must tell that p is a positive integer.

Thus,

=> p = 2, 3, 4, ... & p^2 = 4, 9, 16, ...

Statement 1: Since p^2 is a perfect square number and perfect square numbers have odd numbers of factors, the statement is correct. For example, p^2 = 4; factors of 4 are 4, 2, and 1 (odd numbers of factors). p^2 = 9; factors of 9 are 9, 3, and 1 (odd numbers of factors). This statement is a 'must be true' statement.

Statement 2: Take a couple of examples. p^2 = 4 = 2*2 (two (EVEN) prime factors); p^2 = 9 = 3*3 (two (EVEN) number of positive prime factors); p^2 = 16 = 2*2*2*2 (four (EVEN) number of positive prime factors).

Though p^2 = 16 = 8*2 (a product of a nonprime and a prime number (Odd numbers of prime numbers)), it does not make the statement false since the statement states that p^2 CAN be expressed as the product of an even number of positive prime factors[\b].

The statement does not state that p^2 can ONLY be expressed as the product of an even number of positive prime factors[\b].

Statement 3: For p = 4, the factors are 4, 2 and 1 (Odd numbers of positive factors.) The statement is not a 'must be true' statement.

The correct answer: D

Hope this helps!

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by Ian Stewart » Tue Jul 18, 2017 8:37 pm
Jay@ManhattanReview wrote: I think there is something missing in the question.

We have √p^2 > 1

=> p^2: {2, 3, 4, 5, ...}

=> |p| > 1

=> p < -1 OR p > 1

=> p is not necessarily an integer at all. If it is so, then none of the statements is true.
The question doesn't say that p^2 is an integer greater than 1, which is how I think you've interpreted it; it says that the square root of p^2 is an integer greater than 1, which means p is an integer and p^2 is the square of an integer.

There is one minor issue with the wording - the question talks about "the square root" when it means "the positive square root" (positive numbers have two square roots, one positive, one negative).
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by Matt@VeritasPrep » Sun Jul 23, 2017 5:33 pm
Just a friendly (weekly) PSA to all instructors that ² is much nicer to read than ^2 :)
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