Oct 22 08:00 AM PDT  09:00 AM PDT Join to learn strategies for tackling the longest, wordiest examples of Counting, Sets, & Series GMAT questions Oct 22 09:00 AM PDT  10:00 AM PDT Watch & learn the Do's and Don’ts for your upcoming interview Oct 22 08:00 PM PDT  09:00 PM PDT On Demand for $79. For a score of 4951 (from current actual score of 40+) AllInOne Standard & 700+ Level Questions (150 questions) Oct 23 08:00 AM PDT  09:00 AM PDT Join an exclusive interview with the people behind the test. If you're taking the GMAT, this is a webinar you cannot afford to miss! Oct 26 07:00 AM PDT  09:00 AM PDT Want to score 90 percentile or higher on GMAT CR? Attend this free webinar to learn how to prethink assumptions and solve the most challenging questions in less than 2 minutes. Oct 27 07:00 AM EDT  09:00 AM PDT Exclusive offer! Get 400+ Practice Questions, 25 Video lessons and 6+ Webinars for FREE.
Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 58407

If a and b are both positive integers, is a^2b^2 divisible by 8?
[#permalink]
Show Tags
23 Apr 2015, 03:49
Question Stats:
73% (01:19) correct 27% (01:20) wrong based on 254 sessions
HideShow timer Statistics
If a and b are both positive integers, is \(a^2*b^2\) divisible by 8? (1) a^2 = 4x (2) x is an integer greater than 1 Kudos for a correct solution.
Official Answer and Stats are available only to registered users. Register/ Login.
_________________



Current Student
Joined: 24 Mar 2015
Posts: 35
Concentration: General Management, Marketing
GPA: 3.21
WE: Science (Pharmaceuticals and Biotech)

Re: If a and b are both positive integers, is a^2b^2 divisible by 8?
[#permalink]
Show Tags
23 Apr 2015, 08:38
Bunuel wrote: If a and b are both positive integers, is a^2b^2 divisible by 8?
(1) a^2 = 4x (2) x is an integer greater than 1
Kudos for a correct solution. The original equation can be re written as (a^2b)^2 I believe which would be a^4b (but I think if it is (a^2)*(b^2) the answer will still be the same.). The question is asking us a definite yes or definite no if the original equation is divisible by 8. This will only be solvable if we have the information for both variable a and b. 1) plugging a^2=4x we would get 16x^2b. We still have no information on the new variable X or the original variable B so we do not have sufficient information to answer. Eliminate A and D. 2) This only gives us information about the variable X which is not even in the original equation so this is not sufficient to answer. Eliminate B. Together: we still have no information about variable b so we can not tell what the value of the solved equation will come out to. select answer choice E.
_________________
Please pass me a +KUDOS if you liked my post! Thanks!
Do not compare yourself to others, compare yourself to who you were yesterday.



Retired Moderator
Joined: 06 Jul 2014
Posts: 1220
Location: Ukraine
Concentration: Entrepreneurship, Technology
GMAT 1: 660 Q48 V33 GMAT 2: 740 Q50 V40

Re: If a and b are both positive integers, is a^2b^2 divisible by 8?
[#permalink]
Show Tags
23 Apr 2015, 13:13
funkyleaf wrote: Bunuel wrote: If a and b are both positive integers, is a^2b^2 divisible by 8?
(1) a^2 = 4x (2) x is an integer greater than 1
Kudos for a correct solution. The original equation can be re written as (a^2b)^2 I believe which would be a^4b (but I think if it is (a^2)*(b^2) the answer will still be the same.). The question is asking us a definite yes or definite no if the original equation is divisible by 8. This will only be solvable if we have the information for both variable a and b. 1) plugging a^2=4x we would get 16x^2b. We still have no information on the new variable X or the original variable B so we do not have sufficient information to answer. Eliminate A and D. 2) This only gives us information about the variable X which is not even in the original equation so this is not sufficient to answer. Eliminate B. Together: we still have no information about variable b so we can not tell what the value of the solved equation will come out to. select answer choice E. I am not completely assured but I think if it equal to a^(2b)^2 than it should be equal to a^(4(b^2)) and in this case we can write as (a^2)^(2(b^2)) and as we know from 1 statement that a^2=4x we can rewrite it as (4x)^(2(b^2)) and received 16(x^(2(b^2)) and 16 will be divisible by 8. And 1 statement sufficient. And answer is A But if it \(a^2*b^2\) than answer will be E as you already wrote.
_________________



Current Student
Joined: 24 Mar 2015
Posts: 35
Concentration: General Management, Marketing
GPA: 3.21
WE: Science (Pharmaceuticals and Biotech)

Re: If a and b are both positive integers, is a^2b^2 divisible by 8?
[#permalink]
Show Tags
Updated on: 24 Apr 2015, 09:31
Harley1980 wrote: funkyleaf wrote: Bunuel wrote: If a and b are both positive integers, is a^2b^2 divisible by 8?
(1) a^2 = 4x (2) x is an integer greater than 1
Kudos for a correct solution. The original equation can be re written as (a^2b)^2 I believe which would be a^4b (but I think if it is (a^2)*(b^2) the answer will still be the same.). The question is asking us a definite yes or definite no if the original equation is divisible by 8. This will only be solvable if we have the information for both variable a and b. 1) plugging a^2=4x we would get 16x^2b. We still have no information on the new variable X or the original variable B so we do not have sufficient information to answer. Eliminate A and D. 2) This only gives us information about the variable X which is not even in the original equation so this is not sufficient to answer. Eliminate B. Together: we still have no information about variable b so we can not tell what the value of the solved equation will come out to. select answer choice E. I am not completely assured but I think if it equal to a^(2b)^2 than it should be equal to a^(4(b^2)) and in this case we can write as (a^2)^(2(b^2)) and as we know from 1 statement that a^2=4x we can rewrite it as (4x)^(2(b^2)) and received 16(x^(2(b^2)) and 16 will be divisible by 8. And 1 statement sufficient. And answer is A But if it \(a^2*b^2\) than answer will be E as you already wrote. I still think that if X is an odd number that we multiply 16 by then the whole answer could be nondivisible by 8 therefore my answer remains the same.
_________________
Please pass me a +KUDOS if you liked my post! Thanks!
Do not compare yourself to others, compare yourself to who you were yesterday.
Originally posted by funkyleaf on 23 Apr 2015, 14:06.
Last edited by funkyleaf on 24 Apr 2015, 09:31, edited 1 time in total.



Manager
Joined: 28 Jan 2015
Posts: 124
Concentration: General Management, Entrepreneurship

Re: If a and b are both positive integers, is a^2b^2 divisible by 8?
[#permalink]
Show Tags
23 Apr 2015, 20:33
Bunuel wrote: If a and b are both positive integers, is a^2b^2 divisible by 8?
(1) a^2 = 4x (2) x is an integer greater than 1
Kudos for a correct solution. I think it is E. 1) tells us that a^2 is divisible by at least 4 < not sufficient because even if a^2 = 16, it can be multiplied by any number for b^2 rendering in indivisible by 8. 2) tells us that x can be any number over 1 < still gives us no idea of the value of b



Manager
Joined: 15 May 2014
Posts: 62

Re: If a and b are both positive integers, is a^2b^2 divisible by 8?
[#permalink]
Show Tags
Updated on: 25 Apr 2015, 22:30
a and b are both positive integers is \(a^2\) \(b^2\) divisible by \(8\)?
for the above condition to be true, both \(a\) and \(b\) should be even or, either \(a^2\) or \(b^2\) should be divisible by \(8\)
Statement (1) \(a^2\) = \(4x\); if \(x\) is even, \(a^2\)\(b^2\) is divisible by \(8\) but if \(x\) is odd, \(a^2\)\(b^2\) is NOT divisible by \(8\) Not Sufficient
Statement (2): \(x\) is an integer greater than 1; no info about either \(a\) or \(b\) Not Sufficient
From (1) and (2) No additional information about either \(x\) or \(b\) Not Sufficient
Answer E
Originally posted by sudh on 24 Apr 2015, 04:33.
Last edited by sudh on 25 Apr 2015, 22:30, edited 1 time in total.



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 15309
Location: United States (CA)

Re: If a and b are both positive integers, is a^2b^2 divisible by 8?
[#permalink]
Show Tags
25 Apr 2015, 22:20
Hi Bunuel, The formatting in this question is problematic.... Bunuel wrote: If a and b are both positive integers, is a^2b^2 divisible by 8?
Is "a^2b^2" meant to be interpreted as.... (a^2)(b^2) or (a^2b)^2 or a^[(2b)^2] GMAT assassins aren't born, they're made, Rich
_________________
Contact Rich at: Rich.C@empowergmat.comThe Course Used By GMAT Club Moderators To Earn 750+ souvik101990 Score: 760 Q50 V42 ★★★★★ ENGRTOMBA2018 Score: 750 Q49 V44 ★★★★★



LBS Moderator
Joined: 13 Jan 2015
Posts: 97
Location: United Kingdom
Concentration: Other, General Management

Re: If a and b are both positive integers, is a^2b^2 divisible by 8?
[#permalink]
Show Tags
26 Apr 2015, 02:37
Since i am not sure what the actual question is, i would wait for the solution
Right now, I have two different answers for 2 different prompts..



Math Expert
Joined: 02 Sep 2009
Posts: 58407

Re: If a and b are both positive integers, is a^2b^2 divisible by 8?
[#permalink]
Show Tags
26 Apr 2015, 05:56
EMPOWERgmatRichC wrote: Hi Bunuel, The formatting in this question is problematic.... Bunuel wrote: If a and b are both positive integers, is a^2b^2 divisible by 8?
Is "a^2b^2" meant to be interpreted as.... (a^2)(b^2) or (a^2b)^2 or a^[(2b)^2] GMAT assassins aren't born, they're made, Rich Mathematically a^2b^2 can ONLY mean (a^2)(b^2). If it were something else it would be written differently. P.S. Still edited to avoid ambiguity.
_________________



Math Expert
Joined: 02 Sep 2009
Posts: 58407

Re: If a and b are both positive integers, is a^2b^2 divisible by 8?
[#permalink]
Show Tags
27 Apr 2015, 02:43
Bunuel wrote: If a and b are both positive integers, is \(a^2*b^2\) divisible by 8?
(1) a^2 = 4x (2) x is an integer greater than 1
Kudos for a correct solution. VERITAS PREP OFFICIAL SOLUTION:Solution: E Statement (1) allows us to rewrite the equation as (4x)(b^2). Since we don’t know whether x is an integer, don’t know whether this equation is divisible by 8: if x = 2, for example, then the equation is (8)(b^2), which must be divisible by 8; if x = 2/3, the equation is (8/3)(b^2), which is not necessarily divisible by 8; INSUFFICIENT. Statement (2) on its own is useless – what is x, after all? – so it too is INSUFFICIENT. Together, x > 1, but if x is equal to, say, 9, then 4x(b^2) = 36(b^2), which still isn’t necessarily divisible by 8. (E).
_________________



Current Student
Joined: 12 Aug 2015
Posts: 2562

Re: If a and b are both positive integers, is a^2b^2 divisible by 8?
[#permalink]
Show Tags
24 Aug 2016, 07:41
Amazing Question Here We need to check whether A^2*B^2 is divisible by 8 So we need three two's Statement 1 => x=√4x=> 2√x => x =1 => no x=2 => yes => insuff statement 2 => no clue of A,B=> insuff Combining them => x=2*2 or 2*3 => insuff SMASH THAT E
_________________



Manager
Joined: 20 Apr 2014
Posts: 87

Re: If a and b are both positive integers, is a^2b^2 divisible by 8?
[#permalink]
Show Tags
30 Dec 2016, 04:27
Hi Bunuel I have a question. we know from the question stem that a & b are +ve interger. so a2 must be perfect square since if x = 2 then a2 = 8 and a = 2 square root of 2 which means that a is not integer. therefore combining st.1 and st.2 a bust be 16 at least. Please correct my thinking.



Math Expert
Joined: 02 Sep 2009
Posts: 58407

Re: If a and b are both positive integers, is a^2b^2 divisible by 8?
[#permalink]
Show Tags
30 Dec 2016, 04:38
hatemnag wrote: Hi Bunuel I have a question. we know from the question stem that a & b are +ve interger. so a2 must be perfect square since if x = 2 then a2 = 8 and a = 2 square root of 2 which means that a is not integer. therefore combining st.1 and st.2 a bust be 16 at least. Please correct my thinking. When combining the least value of x can be 4. In this case a^2 = 4*4 = 16 > a = 4. Basically x must also be a perfect square so, for (1)+(2) x can be 4, 9, 16, 25, ...
_________________



Manager
Joined: 20 Apr 2014
Posts: 87

If a and b are both positive integers, is a^2b^2 divisible by 8?
[#permalink]
Show Tags
30 Dec 2016, 05:12
Bunuel wrote: hatemnag wrote: Hi Bunuel I have a question. we know from the question stem that a & b are +ve interger. so a2 must be perfect square since if x = 2 then a2 = 8 and a = 2 square root of 2 which means that a is not integer. therefore combining st.1 and st.2 a bust be 16 at least. Please correct my thinking. When combining the least value of x can be 4. In this case a^2 = 4*4 = 16 > a = 4. Basically x must also be a perfect square so, for (1)+(2) x can be 4, 9, 16, 25, ... OK. so if b=1 and a= 5, 6, 7, 9, 11.. answer is no But if b=1 and a=4, 8, 12, 16, 20 ,... answer if yes So combining is insuff.



Manager
Joined: 25 Mar 2013
Posts: 226
Location: United States
Concentration: Entrepreneurship, Marketing
GPA: 3.5

If a and b are both positive integers, is a^2b^2 divisible by 8?
[#permalink]
Show Tags
23 Jan 2017, 11:42
Is \(\frac{a^2b^2}{8}\) = ? 1, a^2 = 4x 4x b^2/8 = ?? Nothing about b ; Insufficient2, x > 1 Insufficient 3, a^2 = 4x2 = 8 Nothing about b InsufficientE
_________________
I welcome analysis on my posts and kudo +1 if helpful. It helps me to improve my craft.Thank you



NonHuman User
Joined: 09 Sep 2013
Posts: 13384

Re: If a and b are both positive integers, is a^2b^2 divisible by 8?
[#permalink]
Show Tags
25 Jan 2018, 04:06
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). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________




Re: If a and b are both positive integers, is a^2b^2 divisible by 8?
[#permalink]
25 Jan 2018, 04:06






