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M31-05

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M31-05 [#permalink]

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New post 07 Jun 2015, 08:53
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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
[Reveal] Spoiler: OA

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Re M31-05 [#permalink]

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New post 07 Jun 2015, 08:53
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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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Re: M31-05 [#permalink]

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New post 17 Nov 2015, 08:54
Hi Bunuel

Can you please tell what is the difference between this an that problem? I am confused/ Looks like I am missing somethin. Thanks.

If we are told that x is integer, and x^2 is divisible by 27, should not x^2 for sure have 4 factors of 3?

if-n-is-a-positive-integer-and-n-2-is-divisible-by-72-then-129929.html
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Re: M31-05 [#permalink]

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New post 17 Nov 2015, 10:08
shasadou wrote:
Hi Bunuel

Can you please tell what is the difference between this an that problem? I am confused/ Looks like I am missing somethin. Thanks.

If we are told that x is integer, and x^2 is divisible by 27, should not x^2 for sure have 4 factors of 3?

if-n-is-a-positive-integer-and-n-2-is-divisible-by-72-then-129929.html


Doesn't \(3^4*5^4\) have four 3's?
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Re: M31-05 [#permalink]

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New post 16 Jan 2017, 03:03
Bunuel wrote:
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


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)..

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Re: M31-05 [#permalink]

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New post 10 Feb 2017, 02:16
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 ?

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Re: M31-05 [#permalink]

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New post 10 Feb 2017, 02:22

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Re M31-05 [#permalink]

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New post 22 Aug 2017, 14:14
I think this is a high-quality question and I agree with explanation.

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Re M31-05   [#permalink] 22 Aug 2017, 14:14
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