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07 Jun 2015, 08:53
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Difficulty:
55%
(hard)
Question Stats:
58% (01:51) correct
42%
(02:18)
wrong
based on 100
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 factor 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
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
07 Jun 2015, 08:53
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Bookmarks
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 factor 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 the integer \(x\), the minimum value of \(x^2\) that can be a multiple of \(3^3*5^3\) is \(3^4*5^4\). Hence, the smallest possible value for \(x\) 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 factor of every integer \(x\) in T. Only options I and II meet this criterion because \(27 = 3^3\), option III, is not a factor of \(3^2*5^2\).
Answer: C
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 factor 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 the integer \(x\), the minimum value of \(x^2\) that can be a multiple of \(3^3*5^3\) is \(3^4*5^4\). Hence, the smallest possible value for \(x\) 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 factor of every integer \(x\) in T. Only options I and II meet this criterion because \(27 = 3^3\), option III, is not a factor of \(3^2*5^2\).
Answer: C
General Discussion
16 Jan 2017, 03:03
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Bunuel
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
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)..
