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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, 04:49
I hope bunuel shed light on this one..........
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Re: sqrt{ABC} = 504 . Is B divisible by 2? 1) C = 168 2) A is a
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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)
Am I right in this?



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Re: DS Perfect squares
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22 Sep 2013, 12:02
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.



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Re: sqrt{ABC} = 504. Is B divisible by 2?
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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
Hence statements are insufficient. Ans=E



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



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



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Re: sqrt{ABC} = 504. Is B divisible by 2?
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16 May 2016, 22:48
why abc => a*b*c???



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Re: sqrt{ABC} = 504. Is B divisible by 2?
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16 May 2016, 23:14
vishnu440 wrote: why abc => a*b*c??? Hi vishnu440, abc is always a*b*c, but can be easily mistaken as a 3digit number.. A 3digit number will always be mentioned.. say abc is a 3digit number where a, b are ......etc
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Is B divisible by 2?
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24 Aug 2016, 04:11
If \(\sqrt{ABC}=504\) . Is B divisible by 2? (1) C = 168 (2) A is a perfect square
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sqrt{ABC} = 504. Is B divisible by 2?
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Updated on: 25 Aug 2016, 18:27
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 squareIt'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): \(168 = 2^3 \times 3 \times 7 \implies AB = 2^3 \times 3^3 \times 7\) Prove by counterexample \(A = 2^4, B=2^{1} \times 3^3 \times 7 = \text{ odd }\) Anything without a fractional power of 2 in \(B\) would be even. Insufficient
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Originally posted by DAllison2016 on 24 Aug 2016, 05:07.
Last edited by DAllison2016 on 25 Aug 2016, 18:27, edited 1 time in total.



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Re: Is B divisible by 2?
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24 Aug 2016, 05:14
stonecold wrote: If \(\sqrt{ABC}=504\) . Is B divisible by 2?
(1) C = 168 (2) A is a perfect square Given √ABC = 504 We write the factor of 504  2*2*2*7*3*3 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..



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sqrt{ABC} = 504. Is B divisible by 2?
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24 Aug 2016, 05:28
msk0657 wrote: stonecold wrote: If \(\sqrt{ABC}=504\) . Is B divisible by 2?
(1) C = 168 (2) A is a perfect square Given √ABC = 504 We write the factor of 504  2*2*2*7*3*3 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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Re: sqrt{ABC} = 504. Is B divisible by 2?
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24 Aug 2016, 08:17
Hi everyone abhimahna DAllison2016 msk0657here is what i think => The ANSWER here should be C WE aren't told A,B.C are integers. if the Question would mention that A,B,C are integers then ye => C is fine Else here E Gotta be true CC Bunuel
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Re: sqrt{ABC} = 504. Is B divisible by 2?
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27 Aug 2016, 03:36
stonecold wrote: Hi everyone abhimahna DAllison2016 msk0657here is what i think => The ANSWER here should be C WE aren't told A,B.C are integers. if the Question would mention that A,B,C are integers then ye => C is fine Else here E Gotta be true CC BunuelDude, 504 can be written as \(7 * 3^2 * 2^3\) 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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Re: sqrt{ABC} = 504. Is B divisible by 2?
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27 Aug 2017, 00:54
The question is quite simple :
ABC = 504*504 AB = 504*3 We dont know the value of A, therefore the statement A is insufficient
A is a perfect square But this statement doesnt tell us anything about C or B
Combined, AB = 504*3 We can assume A to be any perfect square for example 49, 36, 4, 9 etc.
1. 504*3/49 => Not divisible by 2 2. 504*3/9 => 168 => divisible by 2
Not sufficient E is the answer



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Re: sqrt{ABC} = 504. Is B divisible by 2?
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08 Nov 2018, 22:56
For the option C Let A= 3^2*7^2*2^4 C = 3^2*3^7 In this case B= 3/(7*2) Which is not divisible by 2
I think it is E not C Correct me if I am wrong.




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