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Re: sqrt{ABC} = 504. Is B divisible by 2? [#permalink]
Ravshonbek wrote:
12345678 wrote:
(sqrt)abc=504, is b divisible by 2?

(1) c = 168

(2) a is a perfect square


I get C, combined 1 and 2 sufficient

sqrt:abc=504
or
abc=504*504

now 1: c=168 <=> ab=3*504 so not suff. we need info about a,
now2: a is perfect square, doesnt give anything,
then combine

abc=504^2
168ab=504*504
ab=3*504
ab=9*168 a can be 9, then b=168 or 6 so divisible by 2
or
ab=4*378 so a can be 4 then b is divisible by 2
or
ab=36*42 a can be 36 then b=42 so divisible by 2

C


I think this fine if you assume a b c are all integers. What if a = 81 or 64 ?
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Re: sqrt{ABC} = 504. Is B divisible by 2? [#permalink]
I do agree with Ravshonbek, C

ba=504*3 = 7*2^3*3^3

If I perfect square means just squared number, than even if you take 3^2 or 2^2 there is always left at least one 2 which will be enough to conclude that b is divisible by 2.

Ans: C

And one more thing I think 81 and 64 is not possible to derive.
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Re: sqrt{ABC} = 504. Is B divisible by 2? [#permalink]
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Ferihere wrote:
I do agree with Ravshonbek, C

ba=504*3 = 7*2^3*3^3

If I perfect square means just squared number, than even if you take 3^2 or 2^2 there is always left at least one 2 which will be enough to conclude that b is divisible by 2.

Ans: C

And one more thing I think 81 and 64 is not possible to derive.


I dont understand what you mean by 81 and 64 is not possible to derive.

Stmt 1: c= 168
Stmt2: a is a perfect square

So if abc=504^2 or 254,016
a=81 b=18 2/3 and c=168 would be a valid soloution where b is not divisible by 2.
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sqrt{ABC} = 504. Is B divisible by 2? [#permalink]
\(\sqrt{ABC} = 504\). Is B divisible by 2?

1) C = 168
2) A is a perfect square
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Re: DS- Perfect squares [#permalink]
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]
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]
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]
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]
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]
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.

Consider Kudos if u like this answer
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Re: DS- Perfect squares [#permalink]
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]
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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

Originally posted by EvaJager on 29 Sep 2012, 14:43.
Last edited by EvaJager on 30 Sep 2012, 02:36, 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]
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]
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]
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]
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]
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 \(N = 504.\)

Originally posted by EvaJager on 30 Sep 2012, 02:37.
Last edited by EvaJager on 30 Sep 2012, 04:55, edited 1 time in total.
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