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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.
Re: sqrt{ABC} = 504. Is B divisible by 2?
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19 Sep 2007, 16:53
1
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
Re: sqrt{ABC} = 504. Is B divisible by 2?
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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 ?
Re: sqrt{ABC} = 504. Is B divisible by 2?
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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.
Re: sqrt{ABC} = 504. Is B divisible by 2?
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20 Sep 2007, 11:21
1
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.
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.
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.
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.
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
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.
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
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Updated on: 30 Sep 2012, 02:36
1
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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PhD in Applied Mathematics Love GMAT Quant questions and running.
Re: sqrt{ABC} = 504 . Is B divisible by 2? 1) C = 168 2) A is a
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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.
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.
_________________
If you like my Question/Explanation or the contribution, Kindly appreciate by pressing KUDOS. Kudos always maximizes GMATCLUB worth-Game Theory
If you have any question regarding my post, kindly pm me or else I won't be able to reply
Re: sqrt{ABC} = 504 . Is B divisible by 2? 1) C = 168 2) A is a
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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.
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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PhD in Applied Mathematics Love GMAT Quant questions and running.
Re: sqrt{ABC} = 504 . Is B divisible by 2? 1) C = 168 2) A is a
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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.
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
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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.
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.
_________________
PhD in Applied Mathematics Love GMAT Quant questions and running.
Re: sqrt{ABC} = 504 . Is B divisible by 2? 1) C = 168 2) A is a
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Updated on: 30 Sep 2012, 04:55
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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PhD in Applied Mathematics Love GMAT Quant questions and running.
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