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49% (01:42) correct 51% (01:32) wrong based on 57 sessions
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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
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 \(3^2*5^2\)).
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
beautifully explained just to add we choosed 3^4*5^4[/m] as 3^3*5^3 will have even number of factors (3+1)(3+1)..
Why does the value of x^2 have to be 3^4*5^4? if we assume x^2 is 3^6*5^6 then even 27 will be a divisor or x
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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49% (01:42) correct 
