GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 22 Oct 2019, 00:19 GMAT Club Daily Prep

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

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.  If a and b are both positive integers, is a^2b^2 divisible by 8?

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 58407
If a and b are both positive integers, is a^2b^2 divisible by 8?  [#permalink]

Show Tags

3
3 00:00

Difficulty:   25% (medium)

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.

_________________
Current Student Joined: 24 Mar 2015
Posts: 35
Concentration: General Management, Marketing
GMAT 1: 660 Q44 V38 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

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

1
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
GMAT 1: 660 Q44 V38 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

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 non-divisible 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
GMAT 1: 670 Q44 V38 Re: If a and b are both positive integers, is a^2b^2 divisible by 8?  [#permalink]

Show Tags

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

1
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

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 V
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 15309
Location: United States (CA)
GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: If a and b are both positive integers, is a^2b^2 divisible by 8?  [#permalink]

Show Tags

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
_________________
LBS Moderator B
Joined: 13 Jan 2015
Posts: 97
Location: United Kingdom
Concentration: Other, General Management
Schools: LBS '19 (WL)
GMAT 1: 690 Q48 V36 Re: If a and b are both positive integers, is a^2b^2 divisible by 8?  [#permalink]

Show Tags

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 V
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

1
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 V
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

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 D
Joined: 12 Aug 2015
Posts: 2562
Schools: Boston U '20 (M)
GRE 1: Q169 V154 Re: If a and b are both positive integers, is a^2b^2 divisible by 8?  [#permalink]

Show Tags

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  B
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

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 V
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

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  B
Joined: 20 Apr 2014
Posts: 87
If a and b are both positive integers, is a^2b^2 divisible by 8?  [#permalink]

Show Tags

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  S
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

Is $$\frac{a^2b^2}{8}$$ = ?
1, a^2 = 4x
4x b^2/8 = ??
Nothing about b ; Insufficient
2, x > 1
Insufficient
3, a^2 = 4x2 = 8
Insufficient
E
_________________
I welcome analysis on my posts and kudo +1 if helpful. It helps me to improve my craft.Thank you
Non-Human 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

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
Display posts from previous: Sort by

If a and b are both positive integers, is a^2b^2 divisible by 8?

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne  