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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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01 Oct 2012, 04:58

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)

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: sqrt{ABC} = 504. Is B divisible by 2? [#permalink]

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28 Aug 2014, 22:10

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Re: sqrt{ABC} = 504. Is B divisible by 2? [#permalink]

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28 Aug 2014, 23:44

let us take factors of 504 =2*2*2*7*3*3

LHS has a square root sign. so number of factors double on LHS. so we require 6 2's, 2 7's and 4 3's to equate LHS to RHS.

statement 1 is clearly insufficient. statement 2 also insufficient.

1+2 combined, factors of 168 are 2*2*2*7*3 so balance we require 3 2's, 1 7's and 3 3's to make LHS and RHS equal. a= 2^2*3^2 and b=2*7*3 satisfy above condition and b is a divisible by 2 a=2^4*3^2 and b=(7*3)/2=21/2 also satisfy above condition but b is not divisible by 2

Re: sqrt{ABC} = 504. Is B divisible by 2? [#permalink]

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29 Aug 2014, 00:24

desaichinmay22 wrote:

let us take factors of 504 =2*2*2*7*3*3

LHS has a square root sign. so number of factors double on LHS. so we require 6 2's, 2 7's and 4 3's to equate LHS to RHS.

statement 1 is clearly insufficient. statement 2 also insufficient.

1+2 combined, factors of 168 are 2*2*2*7*3 so balance we require 3 2's, 1 7's and 3 3's to make LHS and RHS equal. a= 2^2*3^2 and b=2*7*3 satisfy above condition and b is a divisible by 2 a=2^4*3^2 and b=(7*3)/2=21/2 also satisfy above condition but b is not divisible by 2

Hence statements are insufficient. Ans=E

Yes, you are right.

But if the question mentioned a, b, c as integers - then answer would be C.

Re: sqrt{ABC} = 504. Is B divisible by 2? [#permalink]

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29 Aug 2014, 01:38

bazc wrote:

desaichinmay22 wrote:

let us take factors of 504 =2*2*2*7*3*3

LHS has a square root sign. so number of factors double on LHS. so we require 6 2's, 2 7's and 4 3's to equate LHS to RHS.

statement 1 is clearly insufficient. statement 2 also insufficient.

1+2 combined, factors of 168 are 2*2*2*7*3 so balance we require 3 2's, 1 7's and 3 3's to make LHS and RHS equal. a= 2^2*3^2 and b=2*7*3 satisfy above condition and b is a divisible by 2 a=2^4*3^2 and b=(7*3)/2=21/2 also satisfy above condition but b is not divisible by 2

Hence statements are insufficient. Ans=E

Yes, you are right.

But if the question mentioned a, b, c as integers - then answer would be C.

That is right. But for this problem, ans has to be E. It is not given that numbers are integers.

Re: sqrt{ABC} = 504. Is B divisible by 2? [#permalink]

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14 May 2016, 12:00

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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abc is always a*b*c, but can be easily mistaken as a 3-digit number.. A 3-digit number will always be mentioned.. say abc is a 3-digit number where a, b are ......etc
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sqrt{ABC} = 504. Is B divisible by 2? [#permalink]

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24 Aug 2016, 05:07

stonecold wrote:

If \(\sqrt{ABC}=504\) . Is B divisible by 2?

504 can be factorised as follows: \(\frac{504}{4} = 126\\1+2+6 = 9 \implies 126 = 9 \times 14\)

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

\(ABC = 2^6 \times 3^4 \times 7^2\)

\(\textbf{(1) } C = 168\)

We know no information about \(B\) or \(A\), so it's only possible to prove that \(B\) is odd. \(B\) would only be odd if \(C\) were divisible by \(2^6 = 64\) 168 is not divisible by 64.

EDIT: stricken through above, we only know one number, even if C were divisible by 64, then \(B=2\) and \(A = \frac{1}{2}\) would allow for an even

Insufficient.

----

(2) A is a perfect square

It's trivial to see that there are multiple combinations of squares, If \(A = 2^{3^2}\), then there would be no twos to distribute. Otherwise a two may or may not be distributed to B to make it even or odd.

Insufficient

(1+2)

From (1): \(168 = 2^3 \times 3 \times 7 \implies AB = 2^3 \times 3^3 \times 7\)

Stat 1: C = 168 then A and B can be 3 or 1. (168*3*1 = 504)...we can't take unique value for A and B.

Stat 2: A is perfect square from the set 2*2*2*7*3*3 , we can take 4 and 9 and also 1..no unique value for A.

Stat 1 + Stat 2: if A is 1 and B = 3 then C = 168. Anyhow 1 is perfect square.

Option C is correct answer..

I doubt on the OA...can you share the full official solution if you have or others can try this..

OA has changed to C now the topics have been merged.

For the OA of E, the numbers do not have to be integers. Let \(A = 144, B = \frac{21}{2}\). B is not even. Let \(A = 4, B = 378\). B is even.
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so, since \(sqrt(ABC) = 504\), we can say ABC = \(7^2 * 3^4 * 2^6\)

Now, Statement 1 has made it clear that C is an integer.

Now, we also know that A*B = \(7 * 3^3 * 2^3\)

We are not sure if B is even. hence, insufficient.

Now Statement B states that A is a perfect Square. No info about B and C, hence insufficient.

Combining,

We know C and product of A and B.

What do you wanna take the value of A. 1/2, if yes, then take it. I know one thing that in order to make the multiplication of A and B an integer, I MUST have 2 in B.

if you wanna take A = 1/7, again take it ( ) , Since I have the product of A and B even, and A is ODD, I can say B MUST be even.

Also, If I go with the normal integral values, then since I have A a perfect square and 3 2's in the product of A and B, I must infer that B must have atleast one 2 in it.

Hence, sufficient.

Do let me know if you have any doubts.
_________________

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