Welcome! Check out our free B-School Guides to learn how you compare with other applicants.
Login or Register
 

Prime & Square

This topic has 24 member replies
Goto page Previous
navalpike Really wants to Beat The GMAT! Default Avatar
Joined
04 May 2009
Posted:
103 messages
Thanked:
5 times
Post Wed Aug 19, 2009 2:47 pm
joyseychow,

Could you kindly post the source of the question?

Need free GMAT or MBA advice from an expert? Register for Beat The GMAT now and post your question in these forums!
joyseychow Really wants to Beat The GMAT! Default Avatar
Joined
15 Dec 2008
Posted:
125 messages
Post 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! Default Avatar
Joined
05 Mar 2011
Posted:
5 messages
Post Mon Jun 27, 2011 6:55 am
navalpike wrote:
My answer is D.

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.

Answer: B

vaibhavdoshi Just gettin' started!
Joined
18 Apr 2011
Posted:
9 messages
Post Mon Jun 27, 2011 9:23 am
IMO B

shingik Just gettin' started!
Joined
19 May 2011
Posted:
28 messages
Post 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! Default Avatar
Joined
23 Sep 2011
Posted:
385 messages
Followed by:
6 members
Thanked:
62 times
Post 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.

user123321

_________________
Just started my preparation Very Happy
Want to do it right the first time.

The Arjun Just gettin' started!
Joined
10 Oct 2010
Posted:
2 messages
Post 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:
Answer shouldn't be D...

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.

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

karthikpandian19 GMAT Titan
Joined
03 Nov 2011
Posted:
1665 messages
Followed by:
63 members
Thanked:
160 times
Target GMAT Score:
750
Post 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?

_________________
Regards,
Karthik
The source of the questions that i post from JUNE 2013 is from KNEWTON

---If you find my post useful, click "Thank" Smile Smile---
---Never stop until cracking GMAT---

eagleeye GMAT Destroyer!
Joined
28 Apr 2012
Posted:
521 messages
Followed by:
48 members
Thanked:
328 times
Test Date:
August 18, 2012
GMAT Score:
770
Post 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 Smile

Thanked by: karthikpandian19
Abhijeet03 Just gettin' started!
Joined
10 Nov 2011
Posted:
5 messages
Thanked:
1 times
Post Tue Nov 06, 2012 3:10 am
Seems u copied answer from your neighbour ..lolz
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.

Best Conversation Starters

1 varun289 34 topics
2 killerdrummer 22 topics
3 sana.noor 21 topics
4 guerrero 14 topics
5 sanaa.rizwan 14 topics
See More Top Beat The GMAT Members...

Most Active Experts

1 image description Brent@GMATPrepNow

GMAT Prep Now Teacher

200 posts
2 image description GMATGuruNY

The Princeton Review Teacher

135 posts
3 image description Jim@StratusPrep

Stratus Prep

106 posts
4 image description Anju@Gurome

Gurome

56 posts
5 image description Jon@Admissionado

Admissionado

44 posts
See More Top Beat The GMAT Experts