Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: sqrt{ABC} = 504. Is B divisible by 2? [#permalink]
19 Sep 2007, 15:53

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? [#permalink]
20 Sep 2007, 07:09

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? [#permalink]
20 Sep 2007, 07:35

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? [#permalink]
20 Sep 2007, 10:21

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.

Re: DS- Perfect squares [#permalink]
30 Jun 2010, 12:06

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.

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.

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.

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.

Re: DS- Perfect squares [#permalink]
14 Oct 2010, 00:18

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

Re: DS- Perfect squares [#permalink]
14 Oct 2010, 00:45

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.

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 _________________

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 [#permalink]
29 Sep 2012, 13:43

1

This post received KUDOS

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 _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Last edited by EvaJager on 30 Sep 2012, 01:36, edited 2 times in total.

Re: sqrt{ABC} = 504 . Is B divisible by 2? 1) C = 168 2) A is a [#permalink]
29 Sep 2012, 17:48

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 [#permalink]
29 Sep 2012, 21:19

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. _________________

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 [#permalink]
30 Sep 2012, 00:04

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.. _________________

Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score !

Re: sqrt{ABC} = 504 . Is B divisible by 2? 1) C = 168 2) A is a [#permalink]
30 Sep 2012, 01:18

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 [#permalink]
30 Sep 2012, 01:37

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.

Answer C

I edited my original post: by mistake it was A = 504 instead of N = 504. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Last edited by EvaJager on 30 Sep 2012, 03:55, edited 1 time in total.

gmatclubot

Re: sqrt{ABC} = 504 . Is B divisible by 2? 1) C = 168 2) A is a
[#permalink]
30 Sep 2012, 01:37

For my Cambridge essay I have to write down by short and long term career objectives as a part of the personal statement. Easy enough I said, done it...