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# M31-05

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Math Expert
Joined: 02 Sep 2009
Posts: 51280
M31-05  [#permalink]

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07 Jun 2015, 07:53
1
4
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Difficulty:

75% (hard)

Question Stats:

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

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Math Expert
Joined: 02 Sep 2009
Posts: 51280
Re M31-05  [#permalink]

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07 Jun 2015, 07:53
2
3
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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17 Nov 2015, 07:54
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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KUDO me plenty

Math Expert
Joined: 02 Sep 2009
Posts: 51280
Re: M31-05  [#permalink]

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17 Nov 2015, 09:08
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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Joined: 22 Nov 2014
Posts: 29
Re: M31-05  [#permalink]

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16 Jan 2017, 02:03
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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Joined: 25 Jan 2015
Posts: 20
Re: M31-05  [#permalink]

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10 Feb 2017, 01:16
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 ?
Math Expert
Joined: 02 Sep 2009
Posts: 51280
Re: M31-05  [#permalink]

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10 Feb 2017, 01:22
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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Posts: 16
Re M31-05  [#permalink]

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22 Aug 2017, 13:14
I think this is a high-quality question and I agree with explanation.
Intern
Joined: 27 May 2017
Posts: 5
GMAT 1: 490 Q39 V19
GPA: 4
Re: M31-05  [#permalink]

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03 Apr 2018, 13:30
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
Math Expert
Joined: 02 Sep 2009
Posts: 51280
Re: M31-05  [#permalink]

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03 Apr 2018, 19:58
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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06 May 2018, 02:23
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
Math Expert
Joined: 02 Sep 2009
Posts: 51280
Re: M31-05  [#permalink]

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06 May 2018, 23:37
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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Joined: 03 Jul 2018
Posts: 1
Re: M31-05  [#permalink]

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24 Sep 2018, 12:39
T=(15, 30, 45,60...............)

Only 15 is the ans. Why are people including x^2 in the set.
Math Expert
Joined: 02 Sep 2009
Posts: 51280
Re: M31-05  [#permalink]

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24 Sep 2018, 19:44
miami009 wrote:
T=(15, 30, 45,60...............)

Only 15 is the ans. Why are people including x^2 in the set.

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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Re: M31-05 &nbs [#permalink] 24 Sep 2018, 19:44
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# M31-05

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