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Fistail, so when the question mentions for 4 different prime numbers, we can actually have 6 prime numbers in total but could be repeated numbers or have exactly 4 prime numbers that are different. so if the question specifies that there are only 4 prime numbers and each is different, then we could answer this with C because we can never have such a number that is a multiple of 4 because we only have 1 even number, which is 2, therefore, C would say that such a combination can not be possible. correct?

Re: If k is a positive integer, is k the square of an integer? [#permalink]
01 Mar 2012, 00:11

6

This post received KUDOS

Expert's post

2

This post was BOOKMARKED

tarek99 wrote:

If k is a positive integer, is k the square of an integer?

(1) k is divisible by 4.

(2) k is divisble by exactly four different prime numbers.

Responding to a pm.

If k is a positive integer, is k the square of an integer?

(1) k is divisible by 4. If \(k=4\) answer is YES, but if \(k=8\) answers is NO. Not sufficient.

(2) k is divisible by exactly four different prime numbers.

We don't know the powers of these primes, so if \(k=2^2*3*5*7\) the answer is NO, but if \(k=(2^2*3*5*7)^2\) the answer is YES (\(k\) equals to square of some integer). Not sufficient.

(1)+(2) Again if \(k=2^2*3*5*7\) the answer is NO, but if \(k=(2^2*3*5*7)^2\) the answer is YES. Not sufficient.

Re: If k is a positive integer, is k the square of an integer? [#permalink]
10 Apr 2012, 10:15

Hello Bunuel,

I chose B and I understand that in order for k to be the square of an integer it's factors need to come in pairs I also understand your explanation. But I'm a little confused, I guess on the wording, that k is divisible by exactly four different prime factors. How can we tell or know not to assume that those four different prime factors aren't squared? I just saw four different and thought there can't be multiples..? Hope this makes sense. Thank you

Re: If k is a positive integer, is k the square of an integer? [#permalink]
10 Apr 2012, 10:29

Expert's post

destroyerofgmat wrote:

Hello Bunuel,

I chose B and I understand that in order for k to be the square of an integer it's factors need to come in pairs I also understand your explanation. But I'm a little confused, I guess on the wording, that k is divisible by exactly four different prime factors. How can we tell or know not to assume that those four different prime factors aren't squared? I just saw four different and thought there can't be multiples..? Hope this makes sense. Thank you

I don't quite understand what you mean by the red part. _________________

Re: If k is a positive integer, is k the square of an integer? [#permalink]
10 Apr 2012, 11:21

Bunuel wrote:

(2) k is divisible by exactly four different prime numbers.

We don't know the powers of these primes, so if \(k=2^2*3*5*7\) the answer is NO, but if \(k=(2^2*3*5*7)^2\) the answer is YES (\(k\) equals to square of some integer). Not sufficient.

This is what I was confused by - so even though you have multiple prime numbers as in (2^2*3*5*7)^2 this is still considered exactly four different prime numbers? Therefore stat 2 is insuff?

Or put differently if I were to ask someone which scenario has exactly four different prime numbers 1 - 2^2*3*5*7 2 - (2^2*3*5*7)^2 the answer would be that they both have four different prime numbers?

Re: If k is a positive integer, is k the square of an integer? [#permalink]
10 Apr 2012, 11:28

1

This post received KUDOS

Expert's post

destroyerofgmat wrote:

Bunuel wrote:

(2) k is divisible by exactly four different prime numbers.

We don't know the powers of these primes, so if \(k=2^2*3*5*7\) the answer is NO, but if \(k=(2^2*3*5*7)^2\) the answer is YES (\(k\) equals to square of some integer). Not sufficient.

This is what I was confused by - so even though you have multiple prime numbers as in (2^2*3*5*7)^2 this is still considered exactly four different prime numbers? Therefore stat 2 is insuff?

Or put differently if I were to ask someone which scenario has exactly four different prime numbers 1 - 2^2*3*5*7 2 - (2^2*3*5*7)^2 the answer would be that they both have four different prime numbers?

Thank you

Yes, both 2^2*3*5*7 and (2^2*3*5*7)^2 have 4 different primes, namely; 2, 3, 5, and 7. How else?

Consider this: 2, 4, 8, and 16 all have only one distinct prime, which is 2.

Re: If k is a positive integer, is k the square of an integer? [#permalink]
10 Apr 2012, 11:30

Exactly 4 different prime numbers means divisible by only 4 different prime numbers and no other factors. But again this is not possible as if you have 4 different prime factors then you can also have non prime factors. Also 1 is a factor of all the numbers. _________________

The question is not can you rise up to iconic! The real question is will you ?

Re: If k is a positive integer, is k the square of an integer? [#permalink]
10 Apr 2012, 11:36

Expert's post

AbhiJ wrote:

Exactly 4 different prime numbers means divisible by only 4 different prime numbers and no other factors. But again this is not possible as if you have 4 different prime factors then you can also have non prime factors. Also 1 is a factor of all the numbers.

That's not correct.

For example: 6 is divisible by exactly two distinct prime factors 2 and 3, but this doesn't mean that 6 divisible by ONLY two factors each of which is a prime. _________________

Re: If k is a positive integer, is k the square of an integer? [#permalink]
10 Apr 2012, 11:56

Expert's post

destroyerofgmat wrote:

Bunuel wrote:

Yes, both 2^2*3*5*7 and (2^2*3*5*7)^2 have 4 different primes, namely; 2, 3, 5, and 7. How else?

Consider this: 2, 4, 8, and 16 all have only one distinct prime, which is 2.

Hope it's clear.

Thanks Bunuel. Another kudos to you, fine sir.

I guess I was thinking that squaring changes TOTAL factors but not different prime factors.

If \(a\) and \(b\) are positive integers then \(a^b\) will have as many different prime factors as \(a\) itself, exponentiation doesn't "produce" primes.

For example: 6, 6^2, 6^3, ..., 6^100 will all have only two distinct primes: 2 and 3. Though total # of factors will naturally be different: # of factors of 6=2*3 is 4, # of factors of 6^2=2^2*3^2 is (2+1)(2+1)=9, ...

Re: If k is a positive integer, is k the square of an integer? [#permalink]
26 Nov 2013, 13:07

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Re: If k is a positive integer, is k the square of an integer? [#permalink]
26 Apr 2014, 04:16

Bunuel wrote:

tarek99 wrote:

If k is a positive integer, is k the square of an integer?

(1) k is divisible by 4.

(2) k is divisble by exactly four different prime numbers.

Responding to a pm.

If k is a positive integer, is k the square of an integer?

(1) k is divisible by 4. If \(k=4\) answer is YES, but if \(k=8\) answers is NO. Not sufficient.

(2) k is divisible by exactly four different prime numbers.

We don't know the powers of these primes, so if \(k=2^2*3*5*7\) the answer is NO, but if \(k=(2^2*3*5*7)^2\) the answer is YES (\(k\) equals to square of some integer). Not sufficient.

(1)+(2) Again if \(k=2^2*3*5*7\) the answer is NO, but if \(k=(2^2*3*5*7)^2\) the answer is YES. Not sufficient.

Answer: E.

Hope it's clear.

Hi Bunnel,

I have a doubt. following are the details

1.As in your previous post you have suggested a perfect square will always have odd number of multiples. Now we have 4 different prime numbers as factor

ex. 2^1 *3^1* 5^1*7^1 so number of factors=( 2*2*2*2) = 16 can not be perfect square we can take diff. powers such as 2,3,4 and verify

Re: If k is a positive integer, is k the square of an integer? [#permalink]
26 Apr 2014, 08:48

Expert's post

PathFinder007 wrote:

Bunuel wrote:

tarek99 wrote:

If k is a positive integer, is k the square of an integer?

(1) k is divisible by 4.

(2) k is divisble by exactly four different prime numbers.

Responding to a pm.

If k is a positive integer, is k the square of an integer?

(1) k is divisible by 4. If \(k=4\) answer is YES, but if \(k=8\) answers is NO. Not sufficient.

(2) k is divisible by exactly four different prime numbers.

We don't know the powers of these primes, so if \(k=2^2*3*5*7\) the answer is NO, but if \(k=(2^2*3*5*7)^2\) the answer is YES (\(k\) equals to square of some integer). Not sufficient.

(1)+(2) Again if \(k=2^2*3*5*7\) the answer is NO, but if \(k=(2^2*3*5*7)^2\) the answer is YES. Not sufficient.

Answer: E.

Hope it's clear.

Hi Bunnel,

I have a doubt. following are the details

1.As in your previous post you have suggested a perfect square will always have odd number of multiples. Now we have 4 different prime numbers as factor

ex. 2^1 *3^1* 5^1*7^1 so number of factors=( 2*2*2*2) = 16 can not be perfect square we can take diff. powers such as 2,3,4 and verify

we can consider this or not?

Please clarify

Thanks.

Firs of all a prefect square has odd number of factors, not multiples, the number of factors is not limited for any integer.

Next, I don't quite understand how are you trying to use that in your solution. Anyway, in my post there are two examples given: one is not a prefect square (\(k=2^2*3*5*7\)) and another is (\(k=(2^2*3*5*7)^2\)). _________________

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