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.

B.

Ignore.

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