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.