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Bunuel
If x is positive, is x prime?

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


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
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Bunuel
If x is positive, is x prime?

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


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

How come statement 1 is not sufficient, when 2 and 3 are both primes?
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ravikumarmishra
Archit3110
Bunuel
If x is positive, is x prime?

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


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

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

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..
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That makes sense. Thanks, I don't know how I missed this scenario, K+1.
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Quote:
How come statement 1 is not sufficient, when 2 and 3 are both primes?

ravikumarmishra

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

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?
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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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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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Thanks arpitkansal, I quite understand now.

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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.
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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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