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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
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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2
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
1
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
Shiridip wrote:
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 ?


x^2 is a perfect square, while 3^4∗5^3 is not. So, x^2 cannot be 3^4∗5^3, it should be 3^4∗5^4.
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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]

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New post 03 Apr 2018, 14:30
Hi Bunuel

I have a doubt here. How come multiple of x^2 is x^4? Shouldn't it be 2*x^2 ?
Please explain this concept to me.
Thank you
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Re: M31-05 [#permalink]

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New post 03 Apr 2018, 20:58
shivangibh wrote:
Hi Bunuel

I have a doubt here. How come multiple of x^2 is x^4? Shouldn't it be 2*x^2 ?
Please explain this concept to me.
Thank you


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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Re: M31-05 [#permalink]

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New post 06 May 2018, 03:23
Hi Bunuel,

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
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Re: M31-05 [#permalink]

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New post 07 May 2018, 00:37
Vinayak Menon wrote:
Hi Bunuel,

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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Re: M31-05   [#permalink] 07 May 2018, 00:37
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