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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]

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19 Sep 2007, 13:46
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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: sqrt{ABC} = 504. Is B divisible by 2? [#permalink]

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19 Sep 2007, 16:53
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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
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Re: sqrt{ABC} = 504. Is B divisible by 2? [#permalink]

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20 Sep 2007, 08:09
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]

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20 Sep 2007, 08:35
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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20 Sep 2007, 11:21
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]

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30 Jun 2010, 12:02
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$$\sqrt{ABC} = 504$$. Is B divisible by 2?

1) C = 168
2) A is a perfect square
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30 Jun 2010, 13:06
Hi,

(1) C = 168

(2) A is a perfect square

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.

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.

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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07 Jul 2010, 06:43
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,

(1) C = 168

(2) A is a perfect square

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.

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.

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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07 Jul 2010, 07:28
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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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07 Jul 2010, 12:41
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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07 Jul 2010, 21:11
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,

(1) C = 168

(2) A is a perfect square

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.

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.

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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14 Oct 2010, 01:18
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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14 Oct 2010, 01:45
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.

Consider Kudos if u like this answer
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29 Sep 2012, 13:56
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.

Let me know if i am wrong.
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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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29 Sep 2012, 14:43
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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.

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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]

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29 Sep 2012, 18:48
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.

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]

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29 Sep 2012, 22:19
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.

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]

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30 Sep 2012, 01:04
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.

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]

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30 Sep 2012, 02:18
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.

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]

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30 Sep 2012, 02:37
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.

I edited my original post: by mistake it was $$A = 504$$ instead of $$N = 504.$$
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Love GMAT Quant questions and running.

Last edited by EvaJager on 30 Sep 2012, 04:55, edited 1 time in total.
Re: sqrt{ABC} = 504 . Is B divisible by 2? 1) C = 168 2) A is a   [#permalink] 30 Sep 2012, 02:37

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