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Set J is comprised of all positive integers x such that x^3 is a multi 31 May 2019, 01:09
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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?

I. 6
II. 9
III. 12

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III
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A
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Set J is comprised of all positive integers x such that x^3 is a multi 31 May 2019, 03:49
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Bunuel
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?

I. 6
II. 9
III. 12

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III
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x^3 = k1(2^3 * 3^2) = k2( 2^3 * 3^3)

so, x will have at least one 2 and one 3.
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Set J is comprised of all positive integers x such that x^3 is a multi 31 May 2019, 07:55
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If x = 6, then x^3 = 6^3 = 216 is divisible by 216 (and automatically by 72 too, because 72 is a factor of 216). So the number 6 is in set J, and clearly then we have numbers in set J which are not divisible by 9 or by 12. So the only possible answer is I only.
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Set J is comprised of all positive integers x such that x^3 is a multi 31 May 2019, 09:42
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Since x is a positive integer, \(x^3\) has to be a perfect cube. 216 is a multiple of 72. So, if \(x^3\) is a multiple of both 72 and 216, it invariably means it’s a multiple of 216.

In general, we can say,
\(x^3\) = 216k, where k is a positive integer.

When we prime factorise 216, we can write it as, 216 = \(2^3\) * \(3^3\). Therefore,

\(x^3\) = \(2^3\) * \(3^3\) * k.

Clearly, 6 has to be a factor of any number \(x^3\), regardless of the value of k. So, 6 also has to be the factor of every member of set J.

9 and 12 may be factors of some elements of the set, which depends on the value of k.

For example, if k = \(2^3\) * \(3^3\), then \(x^3\) = \(2^6\)* \(3^6\), in which case, x = \(2^2\) * \(3^2\). For this value of x, 9 and 12 are factors of x.

But, if k = 1, \(x^3\) = 216, then x = 6. For this value of x, 9 and 12 are not factors of x.

So, the correct answer option is A.

Hope this helps!
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Set J is comprised of all positive integers x such that x^3 is a multi 08 Jun 2024, 08:26
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