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## Prime & Square

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navalpike Really wants to Beat The GMAT!
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Wed Aug 19, 2009 2:47 pm
joyseychow,

Could you kindly post the source of the question?

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joyseychow Really wants to Beat The GMAT!
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Sun Aug 23, 2009 2:56 am
navalpike wrote:
joyseychow,

Could you kindly post the source of the question?
Gmat prep.

DarylB Just gettin' started!
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Mon Jun 27, 2011 6:55 am
navalpike wrote:

A.) I am translating this to mean that if there is �any� prime number in N, then the p^2 version of that prime is also in N. Meaning � Every prime is squared.

If this is the correct interpretation, then this is sufficient. A perfect square, when reduced to primes, will have even numbers as exponents of those primes. For Example.

2*3*5*7 = Not a perfect square

2^2*3^2*5^2*7^2 = perfect square.

Sufficient.
Great interpretation! I think your translation of statement A is correct. But, if a prime factor enters "n" an odd number of times, it may no longer be the square of an interger. Take 8 for example:

8=2*2*2

2 and 2^2=4 are factors, but sqrt(8) is not an interger. Thus A is insufficient.

vaibhavdoshi Just gettin' started!
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Mon Jun 27, 2011 9:23 am
IMO B

shingik Just gettin' started!
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Tue Nov 01, 2011 7:30 pm
There have been examples of proofs that Statement 1 is insufficient (which I think it is). For example someone gave the example of n = 18. So the prime factors of 18 are 2 and 3 right. But 2 squared is NOT a divisor of 18! Does this not disqualify 18 from consideration as a possible n? The same goes for 45 whose prime factors are 3 and 5. 5 squared is not a divisor of 45. However 27 may work because 3 is the only prime factor and 3 squared is a divisor of 27. We know 27 is not a perfect square. Also---maybe the word "distinct" may have made things a bit clearer?

user123321 GMAT Destroyer!
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Wed Nov 02, 2011 4:42 am
is it B?

1 is insufficient because if we consider n as 27, prime factor is 3 and also 27 is divisible by 3^2 but still 27 is not a perfect square.

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The Arjun Just gettin' started!
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Mon Apr 09, 2012 8:09 am
D.

question says that for every prime number p if p is divisor of n then p2 will also be divisor.

n=45, p=3 => n is divisible by p2 but n is not a square of any number.
but we should take p=5 also. In this case p2 is not divisor of n.
So it can be concluded that n is not square of any number.
So first stmt is sufficient.

Second one is known to every one.

Thanks

mehravikas wrote:

Statement 1 -

n = 45, p = 3

p is a divisor of n
also p^2 is a divisor of n

but n in this case is not a square of an integer.

shahdevine wrote:
zeenab wrote:
I think the answer is D as well.

If a number is a perfect square, all of its prime factors have to occur an even number of times in its factorization. Which means, if the number is divisible by a prime number n, it should be divisble by n^2 as well.
So if the corollary holds good as well.
Example -

Take 100 which is 10^2

(5*2)^2 - so the number is divisible by 5 and 5^2 and 2 and 2^2.

you may be right. my interpretation is wrong. we need OAs...

karthikpandian19 GMAT Titan
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Wed Jun 06, 2012 6:54 pm
I got the answer as B

What is the OA?

Can any expert throw light on the Statement 1?

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eagleeye GMAT Destroyer!
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Wed Jun 06, 2012 8:17 pm
OA must be B. There is no other correct answer. Let me explain.

First, there seems to be a debate about "interpretation" of the first statement. There is only one way of doing so.
Hold on, I am going to give my first "verbal" explanation in "quantitative". As Norm Macdonald would say, "Ridiculous". Let`s get on with it.
We are told : "For every prime number p, if p is a divisor of n, then so is p^2
For every prime number p. This only means one thing; it means that p is a prime number.

To interpret it the |other| way, unambiguously, the statement will need to be:
For every prime factor (p) of n, repetitions included, if p is divisor of n, so is p^2. Clearly, this is NOT what the statement asks. So we interpret it as the authors intended. That if there is a prime factor p of n, then there is also p^2 as a factor.

For the question, we are asked, if n is square of a positive integer.
1) Let p = 2 be a factor, then 2^2=4 is a factor, but what if the number was 8. Then it satisfies the 2, 2^2 as factors condition. But 8 is not a square. Hence A is insufficient.

2) We are given that sqrt(n) = m, where (m,n are positive integers);
Squaring both sides, we get n=m^2. Clearly this is what the question asked. Sufficient. Hence B

Let me know if this helps

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Abhijeet03 Just gettin' started!
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Tue Nov 06, 2012 3:10 am
shahdevine wrote:
scoobydooby wrote:
would go with B

stmnt 1 is not sufficient by itself.

if N=9, divisible by 3 and 3^2 . yes N perfect square
if N=27, divisble by 3 and 3^2. no N not perfect square

stmnt 2 sufficient
all perfect squares have integers as their square roots.
cool i'm not crazy...for a minute i was racking my brains. I thought I was from another planet when everyone was getting Ds instead of B.

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