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# M31-05

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Math Expert
Joined: 02 Sep 2009
Posts: 43323

Kudos [?]: 139428 [0], given: 12790

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07 Jun 2015, 07:53
Expert's post
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Difficulty:

75% (hard)

Question Stats:

44% (01:20) correct 56% (01:52) wrong based on 43 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
[Reveal] Spoiler: OA

_________________

Kudos [?]: 139428 [0], given: 12790

Math Expert
Joined: 02 Sep 2009
Posts: 43323

Kudos [?]: 139428 [1], given: 12790

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07 Jun 2015, 07:53
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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 $$3^2*5^2$$).

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Kudos [?]: 139428 [1], given: 12790

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Joined: 12 Aug 2015
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Kudos [?]: 615 [0], given: 1476

Concentration: General Management, Operations
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WE: Management Consulting (Consulting)

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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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Kudos [?]: 615 [0], given: 1476

Math Expert
Joined: 02 Sep 2009
Posts: 43323

Kudos [?]: 139428 [0], given: 12790

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17 Nov 2015, 09:08
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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Kudos [?]: 139428 [0], given: 12790

Intern
Joined: 22 Nov 2014
Posts: 29

Kudos [?]: [0], given: 160

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16 Jan 2017, 02:03
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 $$3^2*5^2$$).

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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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 ?

Kudos [?]: 3 [0], given: 93

Math Expert
Joined: 02 Sep 2009
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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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22 Aug 2017, 13:14
I think this is a high-quality question and I agree with explanation.

Kudos [?]: 5 [0], given: 231

Re M31-05   [#permalink] 22 Aug 2017, 13:14
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# M31-05

Moderators: chetan2u, Bunuel

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