x^(1/3) must definitely be divisible by 4 & 12.mlaboda wrote:Thank you ahead of time for your help!
x is divisible by 144. if x^(1/3) is an integer, then which of the following is x^(1/3) definitely divisible by? (Choose all that apply)
A. 4
B. 8
C. 9
D. 12
Factor Question
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Vemuri
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rossmj
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If x is divisible by 144 then its prime factorization must contain at least (3^2)(2^4). If the cube root of x is an integer then it must contain at least 3 or a multiple of 3 of each of its prime factors. Therefore it must contain at least (3^3)(2^6) the cube root of this leaves us with 3(2^2). Manhattan Gmat does a great job explaining this in its Number Properties book.
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gmat740
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It took me less than 10 seconds to solve this
Only (B) that is 8 has cube-root = 2(integer)
rest don't satisfy this condition
Hope this helps
Now look for the Option which is having an Integral Cube rootif x^(1/3) is an integer
Only (B) that is 8 has cube-root = 2(integer)
rest don't satisfy this condition
Hope this helps
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rossmj
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gmat740, the correct answer is 4 and 12 not 8. After taking the cube root of X we are left with 3(2^2)=12 we can also divide by 4 because we have 2 2s. We want factors of x^1/3. I think you probably read the question a little too quickly but there are 2 answers which is a possiblity mentioned in the question.
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gmat740
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Thanks for correcting me!
I read the question way too fast.
I took x to be multiple of 144 although the question ask you just the reverse.
You see,just a careless mistake can land you up with wrong answer
I read the question way too fast.
I took x to be multiple of 144 although the question ask you just the reverse.
You see,just a careless mistake can land you up with wrong answer
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vittalgmat
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This is a "tongue twister" of a question (maybe brain twister).
4 and 12 for me as well.
Good question.
rgds
4 and 12 for me as well.
Good question.
rgds
