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Updated on: 21 Mar 2013, 04:56
Originally posted by guerrero25 on 21 Mar 2013, 01:09.
Last edited by Bunuel on 21 Mar 2013, 04:56, edited 1 time in total.
Last edited by Bunuel on 21 Mar 2013, 04:56, edited 1 time in total.
Renamed the topic and edited the question.
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
43% (02:03) correct
57%
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wrong
based on 102
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It is known that \(\sqrt[3]{r}\) is a positive integer. Is \(\sqrt[3]{r}\) a prime number?
(1) All the factors of r that are greater than 1 are divisible by 5.
(2) There are exactly four different, positive integers that are factors of r.
(1) All the factors of r that are greater than 1 are divisible by 5.
(2) There are exactly four different, positive integers that are factors of r.
21 Mar 2013, 05:04
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It is known that \(\sqrt[3]{r}\) is a positive integer. Is \(\sqrt[3]{r}\) a prime number?
Given that \(\sqrt[3]{r}=integer\) --> \(r=integer^3\) --> r is a perfect cube.
(1) All the factors of r that are greater than 1 are divisible by 5. If \(r=5^3\), then \(\sqrt[3]{r}=5=prime\) but if \(r=5^6\), then \(\sqrt[3]{r}=25\neq{prime}\). Not sufficient.
(2) There are exactly four different, positive integers that are factors of r. In order r to have 4 factors it must be of the form of \(r=prime^3\) and in this case \(\sqrt[3]{r}=\sqrt[3]{prime^3}=prime\) OR it must be of the form \(r=ab\), where a and b are primes (in this case the number of factors (1+1)(1+1)=4), but in this case r is NOT a perfect suqre, so this case is out. Sufficient.
Answer: B.
THEORY.
Finding the Number of Factors of an Integer
First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.
The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.
Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)
Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.
Hope it helps.
Given that \(\sqrt[3]{r}=integer\) --> \(r=integer^3\) --> r is a perfect cube.
(1) All the factors of r that are greater than 1 are divisible by 5. If \(r=5^3\), then \(\sqrt[3]{r}=5=prime\) but if \(r=5^6\), then \(\sqrt[3]{r}=25\neq{prime}\). Not sufficient.
(2) There are exactly four different, positive integers that are factors of r. In order r to have 4 factors it must be of the form of \(r=prime^3\) and in this case \(\sqrt[3]{r}=\sqrt[3]{prime^3}=prime\) OR it must be of the form \(r=ab\), where a and b are primes (in this case the number of factors (1+1)(1+1)=4), but in this case r is NOT a perfect suqre, so this case is out. Sufficient.
Answer: B.
THEORY.
Finding the Number of Factors of an Integer
First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.
The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.
Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)
Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.
Hope it helps.
21 Mar 2013, 05:09
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guerrero25
We have \(\sqrt[3]{r}\) is an integer.
From F.S 1, for r = \(5^3\), we have \(\sqrt[3]{5^3}\) = 5,a prime. So a YES for the question stem.Again, for r =\(5^6\),we have \(\sqrt[3]{5^6}\) = 25,not a prime. Insufficient.
From F.S 2, we have there are 4 factors for r. Thus, r can only be of the form
r = \(a*b\)or r = \(a^3\), where a,b are primes. For the former one, we wont have \(\sqrt[3]{r}\) as an integer. Thus, only the latter form is valid. Now, for r = \(a^3\), \(\sqrt[3]{r}\) = a, which is a prime. Sufficient.
B.
Ignore.
