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

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

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23 Apr 2015, 03:49
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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.

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

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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.
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Re: If a and b are both positive integers, is a^2b^2 divisible by 8?  [#permalink]

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23 Apr 2015, 13:13
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.

But if it $$a^2*b^2$$ than answer will be E as you already wrote.
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Re: If a and b are both positive integers, is a^2b^2 divisible by 8?  [#permalink]

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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.

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.
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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.
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Re: If a and b are both positive integers, is a^2b^2 divisible by 8?  [#permalink]

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

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Updated on: 25 Apr 2015, 22:30
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.
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Re: If a and b are both positive integers, is a^2b^2 divisible by 8?  [#permalink]

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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]

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

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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..
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Re: If a and b are both positive integers, is a^2b^2 divisible by 8?  [#permalink]

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26 Apr 2015, 05:56
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.
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Re: If a and b are both positive integers, is a^2b^2 divisible by 8?  [#permalink]

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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).
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Re: If a and b are both positive integers, is a^2b^2 divisible by 8?  [#permalink]

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

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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.
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Re: If a and b are both positive integers, is a^2b^2 divisible by 8?  [#permalink]

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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.

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

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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.

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

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23 Jan 2017, 11:42
Is $$\frac{a^2b^2}{8}$$ = ?
1, a^2 = 4x
4x b^2/8 = ??
2, x > 1
Insufficient
3, a^2 = 4x2 = 8
Insufficient
E
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Re: If a and b are both positive integers, is a^2b^2 divisible by 8?  [#permalink]

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