If x is positive, is x prime? (1) x^3 has exactly four distinct posit

Question

If x is positive, is x prime?

(1) x^3 has exactly four distinct positive integer factors.
(2) x^2 - x - 6 = 0

Ypsychotic · Answer

Statement 1) if x is composite, say 10, x^3 will have different distinct positive integer factors. 10^3=(5^3*2^3)=(3+1)*(3+1)=16. It will have exactly four distinct positive integer factors when x is prime and also, it is given x is positive. Hence, Sufficient.

Statement 2) x^2-x-6=0. Breaking it down, (x-3)*(x+2)=0, x=3,-2, but x is positive. So, x=3, hence, prime. Sufficient.

Each is sufficient. IMO, Option D.

Darshi04 · Answer

I think there may be a gap in my conceptual understanding.
 Statement 1:  From what I understand distinct positive integer factors means number of factors of x^3
 Consider 6^3 
6 has 4  distinct  factors, namely 1,2,3, & 6. Cube of 6 will have 1, 2, 3, 8(2^3), 4(2^2), 27(3^3), 9(3^2), 36(6^2), and 6^3, that is 9 distinct factors.
 Is this understanding correct?

Consider 2^3 
2 has 2 distinct factors: 2 & 1.
2^3 will have 8(2^3), 4, 2, &1, that is 4 distinct factors.
2 is a prime number. Hence statement (1) is sufficient.

Unable to understand why OA is B.

Archit3110 · Answer

from 1 
x^3 has 4 distinct factors

so x=2,3 
say 2^3= 8 = 3+1 = 4 factors
or 27= 3^3 = 3+1 = 4 factors

in sufficeint

from 2
x^2 - x - 6 = 0

(x-3)(x+2)
x=3,-2

so 3 is +ve and prime sufficient IMO B

ravikumarmishra · Answer

How come statement 1 is not sufficient, when 2 and 3 are both primes?

Archit3110 · Answer

ravikumarmishra

from 1 x can be (10)^1/3 , so x^3=10^1/3 or say x=10 
also which is 2^1 * 5^1 = 2*2 = 4 factors .. so in sufficient..

ravikumarmishra · Answer

That makes sense. Thanks, I don't know how I missed this scenario, K+1.

vs224 · Answer

ravikumarmishra

from 1 x can be 10 also which is 2^1 * 5^1 = 2*2 = 4 factors .. so in sufficient.. 
--

if x is 10, then x^3 does not have 4 distinct factors. Only prime x^3 has 4 distinct factors. So, how come A is not valid?

Kem12 · Answer

I feel OA is wrong x can't be 10. When x is 10, x^3 will be 1000 and has 16 distinct factors which doesn't satisfy statement 1. Hi  Bunuel , please help clarify if my understanding is correct. Thanks.

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arpitkansal · Answer

IMO If we assume X=(15)^(1/3),then X^3=15=5*3 which also has total of 4 factors. It is not mentioned anywhere that X is a +ve integer.

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Kem12 · Answer

Thanks  arpitkansal , I quite understand now.

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bhaskar123 · Answer

Good question. But I think i can add to the forum. Stmt 1 says X^3 = (some #)^1*(some #)^1 .. or X^3 =(some #)^3 
in both cases total number of factors will be (1+1)*(1+1) = 4 or (3+1) = 4. 
if you choose (3)*(5) then total number of factors is 4 and number is not prime. if you choose 2^3 or 3^3 or 5^3 then number is prime. Hence stmt 1 is not sufficient.

anand0811 · Answer

I think the trick here is X is positive and positive doesn't means that it is an integers.

So X could be (X)^(1/3)
and X3 could have 4 distinct factors.

Therefore It can be 2,3 or 10.

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bumpbot · Answer

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If x is positive, is x prime? (1) x^3 has exactly four distinct posit

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If x is positive, is x prime? (1) x^3 has exactly four distinct posit

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