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If k is a positive integer, is k the square of an integer?  [#permalink]

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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.

Originally posted by tarek99 on 21 Nov 2007, 16:23.
Last edited by Bunuel on 01 Mar 2012, 01:08, edited 1 time in total.
Edited the question and added the OA
Math Expert V
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Re: If k is a positive integer, is k the square of an integer?  [#permalink]

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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.

Hope it's clear.
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Re: DS- is k the square of an integer?  [#permalink]

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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.

B for me.
1) doesnt make any point..can be 4 8 16 20..not suff

2) for a number which is divisible by 4 different prime numbers ,its square cannot be integer.
Manager  Joined: 03 Sep 2006
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B for me

I. k could be 8 or could be 16 - INS
II. K could be 2*3*5*7 or 3*5*7*11 (so do others). Neither number is a perfect square, hence my ans is B
Intern  Joined: 06 Sep 2007
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If k is 2^2*3^2*5^2*7^2 ie (2*3*5*7)^2

on this condition, is B right?
Director  Joined: 03 May 2007
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Schools: University of Chicago, Wharton School
Re: DS- is k the square of an integer?  [#permalink]

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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.

E. K could be (2x3x5x7) or (2x2x3x5x7) or (2x3x5x7)^2.
VP  Joined: 21 Jul 2006
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the OA is E. but i don't understand how. can anyone provide the detailed steps so that we can learn from this? it would be cool for all of us.

thanks
VP  Joined: 21 Jul 2006
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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?
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Re: If k is a positive integer, is k the square of an integer?  [#permalink]

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

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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.

Hope it's clear.
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Re: If k is a positive integer, is k the square of an integer?  [#permalink]

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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.
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Re: If k is a positive integer, is k the square of an integer?  [#permalink]

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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.
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Re: If k is a positive integer, is k the square of an integer?  [#permalink]

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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.
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Re: If k is a positive integer, is k the square of an integer?  [#permalink]

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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, ...

For more on this check Number Theory chapter of Math Book: math-number-theory-88376.html

Hope it helps.
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Re: If k is a positive integer, is k the square of an integer?  [#permalink]

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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.

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?

Thanks.
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If k is a positive integer, is k the square of an integer?  [#permalink]

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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.

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?

Thanks.

Firs of all a prefect square has odd number of factors, not multiples, the number of multiples 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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Re: If k is a positive integer, is k the square of an integer?  [#permalink]

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I failed to interpret the "accurate" meaning of the sentence - "(2) k is divisble by exactly four different prime numbers.". I assumed that there are "only" four divisors of the number - hence marked option C - that is incorrect.
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Re: If k is a positive integer, is k the square of an integer?  [#permalink]

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