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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.

Re: It is known that cuberoot (r) is a positive integer [#permalink]

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21 Mar 2013, 05:09

guerrero25 wrote:

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.

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.

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.

Note the takeaway from this question: All the factors (greater than 1) of a number, n, will not be divisible by the same prime number (other than 1) until and unless the number n is a power of that prime number. This is so because if the number n has 2 prime factors, then there will be factors which have only one of those two primes. If every factor (other than 1) is divisible by the same prime, then the number has only one prime factor.

Ensure that you jot down such logical takeaways when you come across them during practice. Revisiting them again and again will help you build strong fundamentals. You keep joining the dots; the picture will become clear.
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Re: It is known that cuberoot (r) is a positive integer [#permalink]

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08 Sep 2017, 22:37

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