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Set J is comprised of all positive integers x such that x3 i
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22 Oct 2018, 03:56
22 Oct 2018, 03:56
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Question Stats:
38% (01:34) correct
61% (01:43) wrong based on 86 sessions
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Set J is comprised of all positive integers x such that \(x^3\) is a multiple of both 72 and 216. Which of the following integers are factors of every member of set J?
Indicate all such integers.
A. 2
B. 3
C. 6
D. 9
E. 12
Indicate all such integers.
A. 2
B. 3
C. 6
D. 9
E. 12
Set J is comprised of all positive integers x such that x3 i
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16 Mar 2019, 01:39
16 Mar 2019, 01:39
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Set J is comprised of all positive integers x such that x³ is a multiple of both 72 and 216. Which of the following integers are factors of every member of set J?
Indicate all such integers.
A. 2
B. 3
C. 6
D. 9
E. 12
Indicate all such integers.
A. 2
B. 3
C. 6
D. 9
E. 12
Re: Set J is comprised of all positive integers x such that x3 i
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20 Mar 2019, 15:51
20 Mar 2019, 15:51
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We really just need to know the Least Common Multiple of 72 and 216. To do that, we first break them down into their prime factors:
72 = 2*2*2*3*3
216 = 2*2*2*3*3*3
We can see here that 72 is actually a factor of 216 itself. In other words, every number that is a multiple of 72 will also be a multiple of 216. The least common multiple between them is just 216.
So it is possible for x³ to equal 216 itself. We know this because we can write: 216 = (2*3) * (2*3) * (2*3) = 6*6*6 = 6³.
One member of set J, then, is 6. 2, 3, and 6 are factors of 6, but 9 and 12 are not. So A, B, and C are the answers.
72 = 2*2*2*3*3
216 = 2*2*2*3*3*3
We can see here that 72 is actually a factor of 216 itself. In other words, every number that is a multiple of 72 will also be a multiple of 216. The least common multiple between them is just 216.
So it is possible for x³ to equal 216 itself. We know this because we can write: 216 = (2*3) * (2*3) * (2*3) = 6*6*6 = 6³.
One member of set J, then, is 6. 2, 3, and 6 are factors of 6, but 9 and 12 are not. So A, B, and C are the answers.
