Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

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

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

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). _________________