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# M19-17

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Math Expert
Joined: 02 Sep 2009
Posts: 47026

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16 Sep 2014, 01:06
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Difficulty:

65% (hard)

Question Stats:

47% (00:46) correct 53% (00:43) wrong based on 197 sessions

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Is $$p^2$$ a prime number?

(1) $$p \gt 1$$

(2) $$\sqrt{p}$$ is an integer

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Joined: 02 Sep 2009
Posts: 47026

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16 Sep 2014, 01:06
Official Solution:

Statement (1) by itself is insufficient. Consider $$p = 2$$ (the answer is "no") and $$p = \sqrt{2}$$ (the answer is "yes").

Statement (2) by itself is sufficient. If $$\sqrt{p}$$ is an integer, then $$p$$ is also an integer. This means that $$p^2$$ cannot be prime.

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Joined: 28 Aug 2016
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30 Nov 2016, 21:10
Hi Bunuel,

While evaluating the second statement, I considered value of P = 1 , so Sqrt of 1 = 1 and 1 square is not prime, Using both Statement 1 and 2 eliminates the possiblity of P taking the value of 1.

Regards,
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30 Nov 2016, 23:56
akshararaghu wrote:
Hi Bunuel,

While evaluating the second statement, I considered value of P = 1 , so Sqrt of 1 = 1 and 1 square is not prime, Using both Statement 1 and 2 eliminates the possiblity of P taking the value of 1.

Regards,

It's not clear what you are trying to say there. From (2) it follows that p is an integer. The question asks whether p^2 is prime. If p is an integer, then p^2 cannot be a prime. Thus (2) is sufficient on its own (answer B) and we don not need to consider the statements together.
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Joined: 18 Aug 2014
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Location: India
GMAT 1: 740 Q49 V40
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15 Dec 2016, 04:50
even if we consider 1 for the 2nd statement P^2 will not be a prime (1 is not prime). Hence Statement 2 is sufficient
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Joined: 01 Aug 2016
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20 Feb 2017, 22:14
Is p^2 a prime?

1) p>1 means p=2,3,4 --- .. consider p=2 then p^2 is 2^2 is 4.. Not a prime.. it'll work for all the integers greater than 1 so 1st statement is sufficient right.

Am i missing something here? Bunuel
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21 Feb 2017, 00:13
sai897 wrote:
Is p^2 a prime?

1) p>1 means p=2,3,4 --- .. consider p=2 then p^2 is 2^2 is 4.. Not a prime.. it'll work for all the integers greater than 1 so 1st statement is sufficient right.

Am i missing something here? Bunuel

We are not told that p is an integer.

Solution above clearly gives two cases:
Consider $$p = 2$$ (the answer is "no") and $$p = \sqrt{2}$$ (the answer is "yes").
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17 Aug 2017, 05:21
1. $$p=2$$ --> $$p^2=4$$ or$$p=\sqrt{2}$$--> $$p=2$$. Not Sufficient.
2. $$\sqrt{p}=Integer$$. Hence$$p^2$$ has 4 factors for sure (1,$$\sqrt{p}$$,p and $$p^2$$). Hence its not prime Sufficient.
Hence Its B. :)

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M19-17   [#permalink] 17 Aug 2017, 05:21
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# M19-17

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