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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
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
08 Mar 2014, 06:29
Kudos
Bookmarks
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
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.
Answer: D.
Hope it helps.
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
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.
Answer: D.
Hope it helps.
04 Apr 2016, 14:09
Kudos
Bookmarks
Bunuel
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
