Official Solution:


T is the set of all positive integers \(x\) such that \(x^2\) is a multiple of both 27 and 375. Which of the following integers must be a divisor of every integer \(x\) in T?

I. 9

II. 15

III. 27


A. I only
B. II only
C. I and II only
D. I and III only
E. I, II, and III


The least common multiple of \(27 = 3^3\) and \(375 = 3*5^3\) is \(3^3*5^3\).

For an integer \(x\), the least value of \(x^2\) which is a multiple of \(3^3*5^3\) is \(3^4*5^4\), so the least value of \(x\) is \(3^2*5^2\). Therefore, \(T = \{3^2*5^2;\) \(2*(3^2*5^2);\) \(3*(3^2*5^2);\) \(4*(3^2*5^2); ...\}\).

The question asks which of the options must be a divisor of every integer \(x\) in T. So, the answer is I and II only (\(27 = 3^3\) is NOT a divisor of \(3^2*5^2\)).


Answer: C
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Doesn't \(3^4*5^4\) have four 3's?
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Shouldn't it be for the integer X, the least value of X^2 which is a multiple of 3^3∗5^3 is 3^4∗5^3 ?

I think this is a high-quality question and I agree with explanation.

I don't see where the solution says that x^4 is a multiple of x^2 but it's actually true (for integer x).

Integer m is a multiple of integer n means that m/n = integer.

Since x^4/x^2 = x^2 = integer then x^4 is indeed a multiple of x^2. For example, 2^4 = 16 is a multiple of 2^2 = 4.
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The solution does not say that x HAS to be 3^4*5^4.

For an integer \(x\), the least value of \(x^2\) which is a multiple of \(3^3*5^3\) is \(3^4*5^4\), so the least value of \(x\) is \(3^2*5^2\). Therefore, \(T = \{3^2*5^2;\) \(2*(3^2*5^2);\) \(3*(3^2*5^2);\) \(4*(3^2*5^2); ...\}\).

The question asks which of the options must be a divisor of every integer \(x\) in T. So, the answer is I and II only (\(27 = 3^3\) is NOT a divisor of \(3^2*5^2\)).
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This set is not correct. 15^2, 30^2, 45^2, 60^2 are NOT multiples of both 27 and 375.

Correct set is: \(T = \{3^2*5^2;\) \(2*(3^2*5^2);\) \(3*(3^2*5^2);\) \(4*(3^2*5^2); ...\}\).
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