|
Author |
Message |
|
TAGS:
|
|
|
Board of Directors
Joined: 01 Sep 2010
Posts: 3422
|
Re: sqrt{ABC} = 504 . Is B divisible by 2? 1) C = 168 2) A is a [#permalink]
Show Tags
 30 Sep 2012, 04:49
|
|
|
|
|
|
Manager
Joined: 25 Jun 2012
Posts: 64
Location: India
WE: General Management (Energy and Utilities)
|
Re: sqrt{ABC} = 504 . Is B divisible by 2? 1) C = 168 2) A is a [#permalink]
Show Tags
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?
|
|
|
|
|
|
Intern
Joined: 25 Nov 2012
Posts: 12
|
Re: DS- Perfect squares [#permalink]
Show Tags
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.
|
|
|
|
|
|
Manager
Joined: 21 Sep 2012
Posts: 218
Location: United States
Concentration: Finance, Economics
GPA: 4
WE: General Management (Consumer Products)
|
Re: sqrt{ABC} = 504. Is B divisible by 2? [#permalink]
Show Tags
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
|
|
|
|
|
|
Intern
Joined: 27 Aug 2014
Posts: 27
GMAT Date: 09-27-2014
|
Re: sqrt{ABC} = 504. Is B divisible by 2? [#permalink]
Show Tags
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.
|
|
|
|
|
|
Manager
Joined: 21 Sep 2012
Posts: 218
Location: United States
Concentration: Finance, Economics
GPA: 4
WE: General Management (Consumer Products)
|
Re: sqrt{ABC} = 504. Is B divisible by 2? [#permalink]
Show Tags
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.
|
|
|
|
|
|
Intern
Joined: 13 Sep 2015
Posts: 23
Location: India
GPA: 3.2
|
Re: sqrt{ABC} = 504. Is B divisible by 2? [#permalink]
Show Tags
16 May 2016, 22:48
why abc => a*b*c???
|
|
|
|
|
|
Math Expert
Joined: 02 Aug 2009
Posts: 5890
|
Re: sqrt{ABC} = 504. Is B divisible by 2? [#permalink]
Show Tags
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 3-digit number.. A 3-digit number will always be mentioned.. say abc is a 3-digit number where a, b are ......etc
_________________
Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
GMAT online Tutor
|
|
|
|
|
|
BSchool Forum Moderator
Joined: 12 Aug 2015
Posts: 2642
GRE 1: 323 Q169 V154
|
Is B divisible by 2? [#permalink]
Show Tags
24 Aug 2016, 04:11
|
|
|
|
|
|
Manager
Joined: 07 Jul 2016
Posts: 79
GPA: 4
|
sqrt{ABC} = 504. Is B divisible by 2? [#permalink]
Show Tags
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
_________________
Please press +1 Kudos if this post helps.
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.
|
|
|
|
|
|
Retired Moderator
Joined: 26 Nov 2012
Posts: 599
|
Re: Is B divisible by 2? [#permalink]
Show Tags
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..
|
|
|
|
|
|
Manager
Joined: 07 Jul 2016
Posts: 79
GPA: 4
|
sqrt{ABC} = 504. Is B divisible by 2? [#permalink]
Show Tags
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.
_________________
Please press +1 Kudos if this post helps.
|
|
|
|
|
|
BSchool Forum Moderator
Joined: 12 Aug 2015
Posts: 2642
GRE 1: 323 Q169 V154
|
Re: sqrt{ABC} = 504. Is B divisible by 2? [#permalink]
Show Tags
24 Aug 2016, 08:17
|
|
|
|
|
|
Board of Directors
Status: Stepping into my 10 years long dream
Joined: 18 Jul 2015
Posts: 3643
|
Re: sqrt{ABC} = 504. Is B divisible by 2? [#permalink]
Show Tags
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.
_________________
My GMAT Story: From V21 to V40
My MBA Journey: My 10 years long MBA Dream
My Secret Hacks: Best way to use GMATClub
Verbal Resources: All SC Resources at one place | All CR Resources at one place
Blog: Subscribe to Question of the Day Blog
GMAT Club Inbuilt Error Log Functionality - View More.
NEW VISA FORUM - Ask all your Visa Related Questions - here.
How to use an Error Log? Check out my tips! Find a bug in the new email templates and get rewarded with 2 weeks of GMATClub Tests for free
|
|
|
|
|
|
Intern
Joined: 20 Jul 2017
Posts: 4
|
Re: sqrt{ABC} = 504. Is B divisible by 2? [#permalink]
Show Tags
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
|
|
|
|
|
|
|
Re: sqrt{ABC} = 504. Is B divisible by 2?
[#permalink]
27 Aug 2017, 00:54
|
|
|
|
|
Go to page
Previous
1 2
[ 35 posts ]
|
|
|
|
|
|
close
