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sqrt{ABC} = 504. Is B divisible by 2?

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sqrt{ABC} = 504. Is B divisible by 2? [#permalink] New post 30 Jun 2010, 11:02
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\sqrt{ABC} = 504. Is B divisible by 2?

1) C = 168
2) A is a perfect square
[Reveal] Spoiler: OA
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Re: DS- Perfect squares [#permalink] New post 30 Jun 2010, 12:06
Hi,

This is my reply. Hope my answer is clear.

(1) C = 168

(2) A is a perfect square

please explain your answers.

OA soon.

1- C =168
Hence
(2*2*2*3*7)*AB =( 2*2*2*3*3*7)^2
Thus all we know that AB is divisible by 2.
So Answer is not sufficient.

2- A is perfect square
ABC =( 2*2*2*3*3*7)^2
If A = 3*3, C = 7, then B is divisible by 2
If A = (2*2*2)^2 and C = 7, then B is not divisibly by 2.
So Answer is not sufficient.

Taking 1 and 2, we have
(2*2*2*3*7)*AB =( 2*2*2*3*3*7)^2
If B is not divisible by 2, then A = 2*2*2 which is not perfect square.
Hence A cannot be 2*2*2 and hence B is divisible by 2.

Both 1 and 2 are sufficient to answer the question.
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Re: DS- Perfect squares [#permalink] New post 07 Jul 2010, 05:43
I dnt understand,

if squ rt of ABC = 504 doesnt ABC becomes 504*504?

And how did you get =( 2*2*2*3*3*7)^2 for below figure...

(2*2*2*3*7)*AB =( 2*2*2*3*3*7)^2


thanks

jakolik wrote:
Hi,

This is my reply. Hope my answer is clear.

(1) C = 168

(2) A is a perfect square

please explain your answers.

OA soon.

1- C =168
Hence
(2*2*2*3*7)*AB =( 2*2*2*3*3*7)^2
Thus all we know that AB is divisible by 2.
So Answer is not sufficient.

2- A is perfect square
ABC =( 2*2*2*3*3*7)^2
If A = 3*3, C = 7, then B is divisible by 2
If A = (2*2*2)^2 and C = 7, then B is not divisibly by 2.
So Answer is not sufficient.

Taking 1 and 2, we have
(2*2*2*3*7)*AB =( 2*2*2*3*3*7)^2
If B is not divisible by 2, then A = 2*2*2 which is not perfect square.
Hence A cannot be 2*2*2 and hence B is divisible by 2.

Both 1 and 2 are sufficient to answer the question.
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Re: DS- Perfect squares [#permalink] New post 07 Jul 2010, 06:28
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gmatJP wrote:
I dnt understand,

if squ rt of ABC = 504 doesnt ABC becomes 504*504?

And how did you get =( 2*2*2*3*3*7)^2 for below figure...

(2*2*2*3*7)*AB =( 2*2*2*3*3*7)^2


Okay, here's my explanation to this problem. Hope this sheds some light for you.

\sqrt{ABC}=504

Let's prime factorize 504 first.

504 = (2)^3 (3)^2 (7)

So, now we have
\sqrt{ABC}=(2)^3 (3)^2 (7)

To get ABC we square both sides. So you're right in saying that you get 504^2 on the right hand side, but 504^2=((2)^3 (3)^2 (7))^2

So, we have ABC = ((2)^3 (3)^2 (7))^2

Statement 1: C = 168

(AB)168 = ((2)^3 (3)^2 (7))^2


(AB)(2)^3(3)(7) = ((2)^3 (3)^2 (7))^2

So, we know that (AB) = \frac{((2)^3 (3)^2 (7))^2}{(2)^3(3)(7)} This doesn't say much. So it's insufficient.

Statement 2: A is a perfect square.

So, we have ABC = ((2)^3 (3)^2 (7))^2 and A is a perfect square, but then again, this has conflicting values as expressed in the post above.

Statement 1 and 2 together:

(AB)(2)^3(3)(7) = ((2)^3 (3)^2 (7))^2

where A is a perfect square.

Simplifying the equation there gives us (AB) = \frac{((2)^3 (3)^2 (7))^2}{(2)^3(3)(7)} = (2)^3 (7) (3^3)

If A is a perfect square, it either has to be (2)^2 or (3)^2 and for these respective cases, we get B to be
(2) (7) (3^3) or (2)^3 (7) (3). And in both these cases, B is divisible by 2. So C is the solution to the problem.
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Re: DS- Perfect squares [#permalink] New post 07 Jul 2010, 11:41
sqrt{ABC}=504.Is B divisible by 2?


(1) C = 168

(2) A is a perfect square


sqrt{ABC}=504
ABC=504^2

1)
we know C but AB r unknown
hence insuff


2)

A is perfect sq

504
2*252=2*2*126=2*2*2*63=2*2*2*3*3*7

A can be =(2*2*3*3)^2 or (2*2)^2 or (3*3)^2
B and C r still not known
hence insuff

1 and 2
C is known
AB168=504^2
AB=3*504
=3*2*2*2*3*3*7

since A is perf sq
A can be (2*2*3*3)
B will be 3*2*7.it is divisible by 2


id A can be (2*2)
B will be 3*2*7*3*3.it is still divisible by 2

hence C
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Re: DS- Perfect squares [#permalink] New post 07 Jul 2010, 20:11
Yes if sqrt(ABC)= 504, then ABC = 504*504
Let's factorize 504 and get 504 = 2*2*2*3*3*7
For the first condition C=168=2*2*2*3*7
We have ABC= (2*2*2*3*3*7)*(2*2*2*3*3*7)
Thus
(2*2*2*3*7) AB = (2*2*2*3*3*7)*(2*2*2*3*3*7)

Hope it is clear now :)
gmatJP wrote:
I dnt understand,

if squ rt of ABC = 504 doesnt ABC becomes 504*504?

And how did you get =( 2*2*2*3*3*7)^2 for below figure...

(2*2*2*3*7)*AB =( 2*2*2*3*3*7)^2


thanks

jakolik wrote:
Hi,

This is my reply. Hope my answer is clear.

(1) C = 168

(2) A is a perfect square

please explain your answers.

OA soon.

1- C =168
Hence
(2*2*2*3*7)*AB =( 2*2*2*3*3*7)^2
Thus all we know that AB is divisible by 2.
So Answer is not sufficient.

2- A is perfect square
ABC =( 2*2*2*3*3*7)^2
If A = 3*3, C = 7, then B is divisible by 2
If A = (2*2*2)^2 and C = 7, then B is not divisibly by 2.
So Answer is not sufficient.

Taking 1 and 2, we have
(2*2*2*3*7)*AB =( 2*2*2*3*3*7)^2
If B is not divisible by 2, then A = 2*2*2 which is not perfect square.
Hence A cannot be 2*2*2 and hence B is divisible by 2.

Both 1 and 2 are sufficient to answer the question.
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Re: DS- Perfect squares [#permalink] New post 14 Oct 2010, 00:18
anaik100 wrote:
sqrt{ABC}=504.Is B divisible by 2?


(1) C = 168

(2) A is a perfect square


sqrt{ABC}=504
ABC=504^2

1)
we know C but AB r unknown
hence insuff


2)

A is perfect sq

504
2*252=2*2*126=2*2*2*63=2*2*2*3*3*7

A can be =(2*2*3*3)^2 or (2*2)^2 or (3*3)^2
B and C r still not known
hence insuff

1 and 2
C is known
AB168=504^2
AB=3*504
=3*2*2*2*3*3*7

since A is perf sq
A can be (2*2*3*3)
B will be 3*2*7.it is divisible by 2


id A can be (2*2)
B will be 3*2*7*3*3.it is still divisible by 2

hence C

Could you explain why A can only be 2*2*3*3 and 2*2?

In my view:
The problem only states that A is a perfect square
--> A might be 1^2, 2^2, 3^2 4^2 or 5^2
For the first 3 choices of A -> Because B will be 2*... -> B is an even number--> B is divided by 2

For A=4^2 -> B = (3^3 *7)/2 --> (odd * odd * odd * odd)/even --> result is an odd --> undivided by 2

For A=5^2 -> B = (2^3 * 3^3 * 7)/ (5^2) is a fraction b/c the numerator does not include 5 --> undivided by 2

Therefore: If A, B, &C are assumed to be integer -> E is correct

If A, B, C is not integer -> C is correct

Any comment?
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Re: DS- Perfect squares [#permalink] New post 14 Oct 2010, 00:45
sqrt(ABC)=504
ABC=(504)^2=(2^6 )( 3^4)(7^2)
From (1), If C=168, AB= (2^3)(3^3)(7) But B can be anything Odd or even
From (2), A is a perfect square in this case A can be 2^2,3^2,,6^2 But in each case 2 will be left and hence B will be divisible by 2.This employs answer is C.

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Re: DS- Perfect squares [#permalink] New post 29 Sep 2012, 12:56
whiplash2411 wrote:
gmatJP wrote:
I dnt understand,

if squ rt of ABC = 504 doesnt ABC becomes 504*504?

And how did you get =( 2*2*2*3*3*7)^2 for below figure...

(2*2*2*3*7)*AB =( 2*2*2*3*3*7)^2


Okay, here's my explanation to this problem. Hope this sheds some light for you.

\sqrt{ABC}=504

Let's prime factorize 504 first.

504 = (2)^3 (3)^2 (7)

So, now we have
\sqrt{ABC}=(2)^3 (3)^2 (7)

To get ABC we square both sides. So you're right in saying that you get 504^2 on the right hand side, but 504^2=((2)^3 (3)^2 (7))^2

So, we have ABC = ((2)^3 (3)^2 (7))^2

Statement 1: C = 168

(AB)168 = ((2)^3 (3)^2 (7))^2


(AB)(2)^3(3)(7) = ((2)^3 (3)^2 (7))^2

So, we know that (AB) = \frac{((2)^3 (3)^2 (7))^2}{(2)^3(3)(7)} This doesn't say much. So it's insufficient.

Statement 2: A is a perfect square.

So, we have ABC = ((2)^3 (3)^2 (7))^2 and A is a perfect square, but then again, this has conflicting values as expressed in the post above.

Statement 1 and 2 together:

(AB)(2)^3(3)(7) = ((2)^3 (3)^2 (7))^2

where A is a perfect square.

Simplifying the equation there gives us (AB) = \frac{((2)^3 (3)^2 (7))^2}{(2)^3(3)(7)} = (2)^3 (7) (3^3)

If A is a perfect square, it either has to be (2)^2 or (3)^2 and for these respective cases, we get B to be
(2) (7) (3^3) or (2)^3 (7) (3). And in both these cases, B is divisible by 2. So C is the solution to the problem.


Hi Whiplash,

I have a doubt. It is no where given that A is an Integer. So A can be \sqrt{6}^6. In this case B=7 which is not divisible by 2.
Thus Answer E.

Let me know if i am wrong.
Waiting for reply
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Re: sqrt{ABC} = 504 . Is B divisible by 2? 1) C = 168 2) A is a [#permalink] New post 29 Sep 2012, 13:43
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ramana wrote:
\sqrt{ABC} = 504. Is B divisible by 2?

1) C = 168
2) A is a perfect square


(1) 504 = 3 * 168. Therefore, \sqrt{AB\cdot{168}}=3\cdot{168} or AB\cdot{168}=9\cdot{168^2}, from which AB=9\cdot168.
Not sufficient.

(2) Let N be a positive integer such that A=N^2. Then \sqrt{ABC}=N\sqrt{BC}=504.
We can have B=C=1 and N=504 or B=C=2 and N=252.
Not sufficient.

(1) and (2) together:
N\sqrt{B\cdot{168}}=3\cdot168 or N^2B=9\cdot168=8\cdot{27}.
As a perfect square, N^2 can have 2 as a factor necessarily at an even power. Since on the right-hand-side we have a factor of 2^3, B must necessarily have at least one factor of 2.
Sufficient.

Answer C
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Last edited by EvaJager on 30 Sep 2012, 01:36, edited 2 times in total.
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Re: sqrt{ABC} = 504 . Is B divisible by 2? 1) C = 168 2) A is a [#permalink] New post 29 Sep 2012, 17:48
EvaJager wrote:
ramana wrote:
\sqrt{ABC} = 504. Is B divisible by 2?

1) C = 168
2) A is a perfect square


(1) 504 = 3 * 168. Therefore, \sqrt{AB\cdot{168}}=3\cdot{168} or AB\cdot{168}=9\cdot{168^2}, from which AB=9\cdot168.
Not sufficient.

(2) Let N be a positive integer such that A=N^2. Then \sqrt{ABC}=N\sqrt{BC}=504.
We can have B=C=1 and A=504 or B=C=2 and N=252.
Not sufficient.

(1) and (2) together:
N\sqrt{B\cdot{168}}=3\cdot168 or N^2B=9\cdot168=8\cdot{27}.
As a perfect square, N^2 can have 2 as a factor necessarily at an even power. Since on the right-hand-side we have a factor of 2^3, B must necessarily have at least one factor of 2.
Sufficient.

Answer C


Hi Eva,

I have no doubt the answer will be C, if A is an integer. But i want to know where is it mentioned that A is an integer.
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Re: sqrt{ABC} = 504 . Is B divisible by 2? 1) C = 168 2) A is a [#permalink] New post 29 Sep 2012, 21:19
fameatop wrote:
EvaJager wrote:
ramana wrote:
\sqrt{ABC} = 504. Is B divisible by 2?

1) C = 168
2) A is a perfect square


(1) 504 = 3 * 168. Therefore, \sqrt{AB\cdot{168}}=3\cdot{168} or AB\cdot{168}=9\cdot{168^2}, from which AB=9\cdot168.
Not sufficient.

(2) Let N be a positive integer such that A=N^2. Then \sqrt{ABC}=N\sqrt{BC}=504.
We can have B=C=1 and A=504 or B=C=2 and N=252.
Not sufficient.

(1) and (2) together:
N\sqrt{B\cdot{168}}=3\cdot168 or N^2B=9\cdot168=8\cdot{27}.
As a perfect square, N^2 can have 2 as a factor necessarily at an even power. Since on the right-hand-side we have a factor of 2^3, B must necessarily have at least one factor of 2.
Sufficient.

Answer C


Hi Eva,

I have no doubt the answer will be C, if A is an integer. But i want to know where is it mentioned that A is an integer.


(2): A perfect square is an integer.

A and C are integers. Since it isn't mentioned that B is an integer, B=\frac{2^3\cdot{27}}{N^2} is not necessarily an integer, the answer should be indeed E.

For the version assuming that all three numbers, A, B, and C, are integers, the answer is C.
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Re: sqrt{ABC} = 504 . Is B divisible by 2? 1) C = 168 2) A is a [#permalink] New post 30 Sep 2012, 00:04
EvaJager wrote:
ramana wrote:
\sqrt{ABC} = 504. Is B divisible by 2?

1) C = 168
2) A is a perfect square


(1) 504 = 3 * 168. Therefore, \sqrt{AB\cdot{168}}=3\cdot{168} or AB\cdot{168}=9\cdot{168^2}, from which AB=9\cdot168.
Not sufficient.

(2) Let N be a positive integer such that A=N^2. Then \sqrt{ABC}=N\sqrt{BC}=504.
We can have B=C=1 and A=504 or B=C=2 and N=252.
Not sufficient.

(1) and (2) together:
N\sqrt{B\cdot{168}}=3\cdot168 or N^2B=9\cdot168=8\cdot{27}.
As a perfect square, N^2 can have 2 as a factor necessarily at an even power. Since on the right-hand-side we have a factor of 2^3, B must necessarily have at least one factor of 2.
Sufficient.

Answer C


IF C=168...then A*b must be equal to 3?? rite?? then how can A be divisble by 2??

i think statement one is sufficient..answer is no..A can not be divisble by 2...

can any one explain..
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Re: sqrt{ABC} = 504 . Is B divisible by 2? 1) C = 168 2) A is a [#permalink] New post 30 Sep 2012, 01:18
sanjoo wrote:
EvaJager wrote:
ramana wrote:
\sqrt{ABC} = 504. Is B divisible by 2?

1) C = 168
2) A is a perfect square


(1) 504 = 3 * 168. Therefore, \sqrt{AB\cdot{168}}=3\cdot{168} or AB\cdot{168}=9\cdot{168^2}, from which AB=9\cdot168.
Not sufficient.

(2) Let N be a positive integer such that A=N^2. Then \sqrt{ABC}=N\sqrt{BC}=504.
We can have B=C=1 and A=504 or B=C=2 and N=252.
Not sufficient.

(1) and (2) together:
N\sqrt{B\cdot{168}}=3\cdot168 or N^2B=9\cdot168=8\cdot{27}.
As a perfect square, N^2 can have 2 as a factor necessarily at an even power. Since on the right-hand-side we have a factor of 2^3, B must necessarily have at least one factor of 2.
Sufficient.

Answer C


IF C=168...then A*b must be equal to 3?? rite?? then how can A be divisble by 2??

i think statement one is sufficient..answer is no..A can not be divisble by 2...

can any one explain..


NO.
AB=9\cdot168 not 3. See above.
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Re: sqrt{ABC} = 504 . Is B divisible by 2? 1) C = 168 2) A is a [#permalink] New post 30 Sep 2012, 01:37
EvaJager wrote:
ramana wrote:
\sqrt{ABC} = 504. Is B divisible by 2?

1) C = 168
2) A is a perfect square


(1) 504 = 3 * 168. Therefore, \sqrt{AB\cdot{168}}=3\cdot{168} or AB\cdot{168}=9\cdot{168^2}, from which AB=9\cdot168.
Not sufficient.

(2) Let N be a positive integer such that A=N^2. Then \sqrt{ABC}=N\sqrt{BC}=504.
We can have B=C=1 and N=504 or B=C=2 and N=252.
Not sufficient.

(1) and (2) together:
N\sqrt{B\cdot{168}}=3\cdot168 or N^2B=9\cdot168=8\cdot{27}.
As a perfect square, N^2 can have 2 as a factor necessarily at an even power. Since on the right-hand-side we have a factor of 2^3, B must necessarily have at least one factor of 2.
Sufficient.

Answer C



I edited my original post: by mistake it was A = 504 instead of N = 504.
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Last edited by EvaJager on 30 Sep 2012, 03:55, edited 1 time in total.
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Re: sqrt{ABC} = 504 . Is B divisible by 2? 1) C = 168 2) A is a [#permalink] New post 30 Sep 2012, 03:49
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Re: sqrt{ABC} = 504 . Is B divisible by 2? 1) C = 168 2) A is a [#permalink] New post 01 Oct 2012, 03:58
I dont know whether my apporach is right or not..

sqrt{ABC}= 504

here, LHS is sqrt and RHS is integer so LHS must be a n integer value when sqrt is removed.

statement 1 : C=168, nothinh about A and B so insufficient.
statement 2 : A is a perfect square

now A is pefect square, we have value of C=168.

so to make it of even number of power, suppose, B=C=168.

so A,B and C all are of even power, so both the statement are together suffice.B=168 is divisible by 2.
(question stem doesnot mention that all the three are distinct numbers. so we can assume B=C)

Am I right in this?
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Re: DS- Perfect squares [#permalink] New post 22 Sep 2013, 11:02
anaik100 wrote:
sqrt{ABC}=504.Is B divisible by 2?


(1) C = 168

(2) A is a perfect square


sqrt{ABC}=504
ABC=504^2

1)
we know C but AB r unknown
hence insuff


2)

A is perfect sq

504
2*252=2*2*126=2*2*2*63=2*2*2*3*3*7

A can be =(2*2*3*3)^2 or (2*2)^2 or (3*3)^2
B and C r still not known
hence insuff

1 and 2
C is known
AB168=504^2
AB=3*504
=3*2*2*2*3*3*7

since A is perf sq
A can be (2*2*3*3)
B will be 3*2*7.it is divisible by 2


id A can be (2*2)
B will be 3*2*7*3*3.it is still divisible by 2

hence C





Answer C is based on the assumption that B is an integer, and no such information is given anywhere in the question.
I believe, answer C wont be valid in case, c=168, a=504^2, B=1/168.
Kindly correct if I am wrong. Thanks.
Re: DS- Perfect squares   [#permalink] 22 Sep 2013, 11:02
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(sqrt)abc=504, is b divisible by 2? 12345678 5 19 Sep 2007, 12:46
if A and B are positive integers, is $$a^2 + b^2$$ divisable FN 15 03 Aug 2007, 06:42
sqrt(abc) = 504. Is B divisible by 2? (1) C = 168 (2) A is a Himalayan 3 16 Jun 2007, 13:35
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sqrt{ABC} = 504. Is B divisible by 2?

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