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Prime & Square

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by joyseychow » Thu Aug 06, 2009 6:35 am
Is the positive integer n equal to the square of an integer?
(1) For every prime number p, if p is a divisor of n, then so is p2.
(2) sqrt root (n) is an integer.

I got this correct. But I'm still not sure about Stmt 1. Pls. Explain. Thks!
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Source: — Data Sufficiency |
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Re: Prime & Square

by shahdevine » Thu Aug 06, 2009 6:53 am
joyseychow wrote:Is the positive integer n equal to the square of an integer?
(1) For every prime number p, if p is a divisor of n, then so is p2.
(2) sqrt root (n) is an integer.

I got this correct. But I'm still not sure about Stmt 1. Pls. Explain. Thks!

statement 1)

restate:

for every prime number that is a factor of n then p^2 is also a factor of n.

let's make our prime number(p) = 5. N at least would have to equal 25 in order for p^2 to be a factor of N. So n=p^2. However, n could have additional primes that are not p. For instance n could equal 50, whose factors are 5^2 and 2. If this were the case n does not equal the square of an integer. It equals the square of an integer times another prime number. So insufficient.

Statement 2)

sqroot n = an integer
n = integer squared

sufficient

B is answer.

pls confirm.

you got this!
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by navalpike » Thu Aug 06, 2009 8:16 am
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.
Last edited by navalpike on Fri Aug 07, 2009 9:04 am, edited 1 time in total.
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by zeenab » Thu Aug 06, 2009 9:18 am
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?
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by shahdevine » Thu Aug 06, 2009 3:54 pm
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...
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by mehravikas » Thu Aug 06, 2009 5:55 pm
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...
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Re: Prime & Square

by tohellandback » Thu Aug 06, 2009 6:15 pm
joyseychow wrote:Is the positive integer n equal to the square of an integer?
(1) For every prime number p, if p is a divisor of n, then so is p2.
(2) sqrt root (n) is an integer.

I got this correct. But I'm still not sure about Stmt 1. Pls. Explain. Thks!
edited..
IMO B- look for scoobydooby's example
1) it says that all the prime factors of n have even powers- i was wrong
it says that the powers are greater than 2. Can be even. Can be odd
Not sufficient

2) Sufficient

@mehravikas,
1) says every prime number.
In your example 45= 3^2*5.
you haven't considered 5.
Last edited by tohellandback on Thu Aug 06, 2009 10:57 pm, edited 1 time in total.
The powers of two are bloody impolite!!
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by scoobydooby » Thu Aug 06, 2009 10:50 pm
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.
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by shahdevine » Sun Aug 09, 2009 6:57 am
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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by arorag » Sun Aug 09, 2009 7:07 am
should be B
For A lets have n=8, p= 2, p2= 4
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by joyseychow » Wed Aug 12, 2009 9:36 pm
Yes OA is indeed B
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by vittalgmat » Wed Aug 12, 2009 10:30 pm
The question is a bit confusing.. esp stmt 1.

But this is a good discussion.
I got B with my interpretation.


thanks everyone.
-V
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by vikram_k51 » Thu Aug 13, 2009 10:14 am
Is the positive integer n equal to the square of an integer?
(1) For every prime number p, if p is a divisor of n, then so is p2.
(2) sqrt root (n) is an integer.

D
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by dikku07 » Thu Aug 13, 2009 1:18 pm
in favour of B
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by jjk » Mon Aug 17, 2009 6:47 pm
B

You can test statement 1 by setting n equal to 9 or 18 and p equal to 3.

If n = 9 and p = 3, 9/3 and 9/9 yield integers.

If n = 18 and p = 3, 18/3 and 18/9 yield integers.

However, 18 is NOT a square of an integer. Statement 1 is thus insufficient.
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