joyseychow,

Could you kindly post the source of the question?

navalpike wrote:joyseychow,

Could you kindly post the source of the question?

Gmat prep.

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

IMO B

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?

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

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...

I got the answer as B

What is the OA?

Can any expert throw light on the Statement 1?

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

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.

1) Try n=72: n has 2 prime factors 2 and 3. n satisfies the stat 1 but not being a square of an integer.
=> insufficient

2) Clearly sufficient

Choose B

According to me,Answer should be B

Is the positive integer n equal to the square of an integer?

Statement 1:

For every prime number p, if p is a divisor of n, then so is p^2.

Let say p = 5

P can be divisor of 50 or of 225

and p^2 also divides both. So Not Sufficient as 50 is not a perfect square and 225 is.

Statement 2:
sqrt(n) is an integer.

Which itself explains that positive integer n is a square of an integer.

Also, OA is mentioned here:

https://gmatclub.com/forum/is-the-positi ... 17148.html

It is B..

Thanks

Let's settle this one!

Is n = k², where k is an integer?

S1:: For EVERY prime number p, if p is a divisor of n, then so is p².

This means that n has AT LEAST TWO of each of its unique prime factors. For instance, if 3 is a divisor of n, then 3² is a divisor of n; if n is a divisor of n, then 2² is a divisor of n, etc. So the example of n = 18 given before isn't valid: 2 divides n, so 2² must divide n - we can't pick a number that violates the property given. We need to pick numbers that have AT LEAST TWO of each unique prime factor.

That said, here are two (contradictory) integers that would satisfy S1. Suppose n = 36. Then n has AT LEAST TWO of each of its prime factors (2 and 3) and IS a perfect square. Now suppose that n = 72. n again has AT LEAST two of each of its prime factors (2 and 3), but is NOT a perfect square, as n has an odd number of 2s in its prime factorization. INSUFFICIENT.

The key distinction here is that saying a number has AT LEAST TWO of each prime factor is not the same thing as saying EXACTLY TWO of each prime factor. To be a perfect square, a number must have an EVEN number of each of its unique primes, as you must be able to group the unique primes into TWO identical roots. To illustrate:

36 = 2*2*3*3 = (2*3) * (2*3) = a perfect square
72 = 2*2*2*3*3 = (2*3) * (2*3) * 2 = not a perfect square

S2:: √n is an integer. Then n = √n * √n = integer * itself = a perfect square. SUFFICIENT!

IMO:B

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) √n is an integer.

Question stem, rephrased:
Is n a perfect square?

Statement 1: For every prime number p, if p is a divisor of n, then so is p².
Let p=2, implying that both p=2 and p²=4 are factors of n.
It's possible that n=4, in which case n is a perfect square.
It's possible that n=8, in which case n is NOT a perfect square.
INSUFFICIENT.

Statement 2: √n is an integer.
In other words:
√n = k, where k is an integer.
Squaring both sides, we get:
n = k².
Thus, n is a perfect square.
SUFFICIENT.

The correct answer is B.