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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If the square root of p^2 is an integer greater than 1
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GMATsid2016
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Hi GMATsid2016,
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
GMAT assassins aren't born, they're made,
Rich
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
GMAT assassins aren't born, they're made,
Rich
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Matt@VeritasPrep
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I and III can't both be true - it's one or the other - so try a number to see.
Suppose p = 4. Then p² = 16, and it has five factors: 1, 2, 4, 8, and 16. So I is in, III is out.
From there, let's see if II works. 2 * 2 * 2 * 2 = 16, OK, but maybe we were just lucky. How about p = 5 and p² = 25? Works again: 5 * 5 = 25.
So if we had to guess, we'd guess D.
Suppose p = 4. Then p² = 16, and it has five factors: 1, 2, 4, 8, and 16. So I is in, III is out.
From there, let's see if II works. 2 * 2 * 2 * 2 = 16, OK, but maybe we were just lucky. How about p = 5 and p² = 25? Works again: 5 * 5 = 25.
So if we had to guess, we'd guess D.
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Matt@VeritasPrep
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If we had more time (or were more anxious), we could try to show why II is true.
If p is prime, then II is obviously true, since p * p = p², and p² is the product of two primes.
If p ISN'T prime, then let's say p = (a� * a₂ * ... * an), where all the a's are primes.
Since p * p =
(a� * a₂ * ... * an) * (a� * a₂ * ... * an) =
(a�)² * (a₂)² * ... * (an)² =
two a1s * two a2s * ... * two ans
we again have the product of an even number of primes, so we're good to go!
If p is prime, then II is obviously true, since p * p = p², and p² is the product of two primes.
If p ISN'T prime, then let's say p = (a� * a₂ * ... * an), where all the a's are primes.
Since p * p =
(a� * a₂ * ... * an) * (a� * a₂ * ... * an) =
(a�)² * (a₂)² * ... * (an)² =
two a1s * two a2s * ... * two ans
we again have the product of an even number of primes, so we're good to go!
