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# T is the set of all positive integers x such that x^2 is a multiple of

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Math Expert
Joined: 02 Sep 2009
Posts: 58381
T is the set of all positive integers x such that x^2 is a multiple of  [#permalink]

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12 May 2015, 04:45
1
9
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Difficulty:

85% (hard)

Question Stats:

49% (02:15) correct 51% (02:14) wrong based on 204 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

Kudos for a correct solution.

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Posts: 58381
Re: T is the set of all positive integers x such that x^2 is a multiple of  [#permalink]

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18 May 2015, 06:02
2
5
Bunuel wrote:
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

Kudos for a correct solution.

OFFICIAL SOLUTION:

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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Re: T is the set of all positive integers x such that x^2 is a multiple of  [#permalink]

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12 May 2015, 09:40
1
x = sq rt 27*375 = 3^2*5*sqrt5

I 3^2 - yes, its a factor
II 3*5 - yes, its a factor
III 3^3 - not a factor

Hence C
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T is the set of all positive integers x such that x^2 is a multiple of  [#permalink]

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12 May 2015, 21:08
2
Bunuel wrote:
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

Ans: C
Solution:
T->{x} where x^2 is a multiple of both 27 and 375 means 3^3 and (5^3)*3
means x must contain 3^2 and 5^2
so with these conditions we know that 9=3^2 and 15=3*5 both have required factors for the divisibility of lowest int for x which is 9*25
but 27 is not a divisor because it can't divide 9*25 fully.

so Ans : C
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Re: T is the set of all positive integers x such that x^2 is a multiple of  [#permalink]

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08 Mar 2017, 02:54
2
27 = 3^3
375 = 3 x 5^3
LCM = 3^3 x 5^3

if x^2 is divisible by LCM, then x^2 should have 3^4 x 5^4
therefore x should have 3^2 x 5^2 = 225. therefore every integer in T will be divisible by 225 or factors of 225 which is 1, 3,5,9,15, 25,45, 75, 225.

looking at the option we can see that Option C is correct
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Re: T is the set of all positive integers x such that x^2 is a multiple of  [#permalink]

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10 Mar 2017, 10:35
1
Bunuel wrote:
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

Since x^2 is a multiple of both 27 and 375, x^2 is a multiple of the LCM of 27 and 375.

27 = 3^3

375 = 25 x 15 = 5^3 x 3^1

Thus, the LCM of 27 and 375 is 3^3 x 5^3 = 15^3 and x^2 must be a multiple of 15^3. Since x^2 is a perfect square and a multiple of 15^3, the minimum value of x^2 must be be 15 * 15^3 = 15^4. Since x is the square root of x^2, taking the square root of 15^4, the minimum value of x must be 15^2 = 225. We see that both 9 and 15 are factors of 225, but 27 is not.

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Re: T is the set of all positive integers x such that x^2 is a multiple of  [#permalink]

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01 Jul 2019, 08:22
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Re: T is the set of all positive integers x such that x^2 is a multiple of   [#permalink] 01 Jul 2019, 08:22
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