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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, 02:49
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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, 07:38
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, 12:13
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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.
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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, 13:06
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.
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Last edited by funkyleaf on 24 Apr 2015, 08:31, 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, 19:33
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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24 Apr 2015, 03:33
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
Last edited by sudh on 25 Apr 2015, 21:30, 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, 21:20
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
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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, 01:37
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, 04:56



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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, 01:43
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, 06:41
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, 03:27
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.



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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, 03:38



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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, 04:12
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.



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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, 10:42
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
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