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If A and B are positive integers and A^3 is divisible by 24

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If A and B are positive integers and A^3 is divisible by 24  [#permalink]

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New post Updated on: 05 Mar 2018, 06:53
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If A and B are positive integers and \(A^3\) is divisible by 24, then is \(AB^2\) a multiple of 216?

(1) B is a multiple of 6

(2) B is divisible by 30

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Originally posted by itisSheldon on 05 Mar 2018, 04:49.
Last edited by itisSheldon on 05 Mar 2018, 06:53, edited 1 time in total.
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Re: If A and B are positive integers and A^3 is divisible by 24  [#permalink]

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New post 05 Mar 2018, 05:09
itisSheldon wrote:
If A and B are positive integers and A^3\(\) is divisible by 24, then is AB^2\(\) a multiple of 216?

(1) B is a multiple of 6

(2) B is divisible by 30



\(A^3\) div by 24 (means div by \(24=2^3*3\))...
so A should be div by 2*3 or 6

now \(216 = 2^3*3^3\)...
since A is div by 6, B^2 must be div by \(2^2*3^2\) or B must be div by \(\sqrt{2^2*3^2}=6\) for AB^2 to be div by 216

lets see the statements..

(1) B is a multiple of 6
so suff

(2) B is divisible by 30
so B will be div by 6
suff

D
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: If A and B are positive integers and A^3 is divisible by 24  [#permalink]

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New post 05 Mar 2018, 07:01
chetan2u wrote:
itisSheldon wrote:
If A and B are positive integers and A^3\(\) is divisible by 24, then is AB^2\(\) a multiple of 216?

(1) B is a multiple of 6

(2) B is divisible by 30



\(A^3\) div by 24 (means div by \(24=2^3*3\))...
so A should be div by 2*3 or 6

now \(216 = 2^3*3^3\)...
since A is div by 6, B^2 must be div by \(2^2*3^2\) or B must be div by \(\sqrt{2^2*3^2}=6\) for AB^2 to be div by 216

lets see the statements..

(1) B is a multiple of 6
so suff

(2) B is divisible by 30
so B will be div by 6
suff

D



Thank you for the explanation!

My understanding was if A^3 is divisible by 24, then A^3 must be equal to 24k (multiple of k) and therefore A "must be divisible" by 2*(3^1/3)... Please help me identify where am I going wrong.
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Re: If A and B are positive integers and A^3 is divisible by 24  [#permalink]

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New post 05 Mar 2018, 07:12
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1
urvashis09 wrote:
chetan2u wrote:
itisSheldon wrote:
If A and B are positive integers and A^3\(\) is divisible by 24, then is AB^2\(\) a multiple of 216?

(1) B is a multiple of 6

(2) B is divisible by 30



\(A^3\) div by 24 (means div by \(24=2^3*3\))...
so A should be div by 2*3 or 6

now \(216 = 2^3*3^3\)...
since A is div by 6, B^2 must be div by \(2^2*3^2\) or B must be div by \(\sqrt{2^2*3^2}=6\) for AB^2 to be div by 216

lets see the statements..

(1) B is a multiple of 6
so suff

(2) B is divisible by 30
so B will be div by 6
suff

D



Thank you for the explanation!

My understanding was if A^3 is divisible by 24, then A^3 must be equal to 24k (multiple of k) and therefore A "must be divisible" by 2*(3^1/3)... Please help me identify where am I going wrong.



pretty much in the same line BUT A is an integer so it cannot have a value of 3rd root of 3 and minimum value has to be 2*3=6
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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If A and B are positive integers and A^3 is divisible by 24  [#permalink]

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New post 05 Mar 2018, 07:20
urvashis09 wrote:
chetan2u wrote:
itisSheldon wrote:
If A and B are positive integers and A^3\(\) is divisible by 24, then is AB^2\(\) a multiple of 216?

(1) B is a multiple of 6

(2) B is divisible by 30



\(A^3\) div by 24 (means div by \(24=2^3*3\))...
so A should be div by 2*3 or 6

now \(216 = 2^3*3^3\)...
since A is div by 6, B^2 must be div by \(2^2*3^2\) or B must be div by \(\sqrt{2^2*3^2}=6\) for AB^2 to be div by 216

lets see the statements..

(1) B is a multiple of 6
so suff

(2) B is divisible by 30
so B will be div by 6
suff

D



Thank you for the explanation!

My understanding was if A^3 is divisible by 24, then A^3 must be equal to 24k (multiple of k) and therefore A "must be divisible" by 2*(3^1/3)... Please help me identify where am I going wrong.



Please note that if A^3 is divisible by 24(i.e Factors of A^3 are 2 and 3) the A must have 2 and 3 as their factors. For instance if A^3 is divisible by 343 then A must be divisible by 7. Similarly if A^3 is divisible by 18 then A must be divisible by 2*3.
Hope it helps and correct me if i am wrong.

--== Message from the GMAT Club Team ==--

THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION.
This discussion does not meet community quality standards. It has been retired.


If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.

_________________

Please mention my name in your valuable replies.

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If A and B are positive integers and A^3 is divisible by 24 &nbs [#permalink] 05 Mar 2018, 07:20
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If A and B are positive integers and A^3 is divisible by 24

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