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x^2 is divisible by both 40 and 75. If x has exactly three distinct

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x^2 is divisible by both 40 and 75. If x has exactly three distinct  [#permalink]

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New post 22 May 2019, 23:29
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A
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C
D
E

Difficulty:

  45% (medium)

Question Stats:

65% (02:25) correct 35% (02:20) wrong based on 55 sessions

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Re: x^2 is divisible by both 40 and 75. If x has exactly three distinct  [#permalink]

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New post 22 May 2019, 23:46
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Bunuel wrote:
x is a positive integer, and \(x^2\) is divisible by both 40 and 75. If x has exactly three distinct prime factors, which of the following could be the value of x?

I. 60
II. 200
III. 240

A. I only
B. II only
C. III only
D. I and III only
E. I, II and III


X can't be 200 because 200 doesn't have 3 distinct prime numbers.
200= 2^3 * 5^2

Now For a number to be divisible by 40 it must have (2^3 & 5)
Now For a number to be divisible by 75 it must have (3 & 5^2)

let's check 60, 3 distinct prime factors.
60*60= 2^4 * 3^2 *5^2 Div. both by 40 and 75

let's check 240, 3 distinct prime factors.
240*240= 2^8 * 3^2 * 5^2.

So answer I & III only.
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Re: x^2 is divisible by both 40 and 75. If x has exactly three distinct  [#permalink]

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New post 23 May 2019, 09:49
Bunuel wrote:
x is a positive integer, and \(x^2\) is divisible by both 40 and 75. If x has exactly three distinct prime factors, which of the following could be the value of x?

I. 60
II. 200
III. 240

A. I only
B. II only
C. III only
D. I and III only
E. I, II and III


use answer options to solve the question
60 & 240 have distinct 3 prime factors and are divisible by 40 & 75
IMO D
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Re: x^2 is divisible by both 40 and 75. If x has exactly three distinct  [#permalink]

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New post 28 May 2019, 20:57
This is a good question which tests you on concepts of factors; it also tests your knowledge of perfect squares and LCM to some extent.

Since x is a positive integer, \(x^2\) will be a perfect square. The question also says that \(x^2\) is divisible by both 40 and 75. So, we can safely say that \(x^2\) should be divisible by the LCM of 40 and 75. Let us look at both these numbers in bit more detail.

40 = \(2^3\) * 5

75 = \(5^2\) * 3.

Therefore, LCM of 40 and 75 = \(2^3\) * 3 * \(5^2\) = 600. Since 600 is not a perfect square, \(x^2\) cannot be equated to 600. We can say that \(x^2\) may be written in the form of 600k where k is a positive integer.

The smallest integer by which 600 should be multiplied to make it a perfect square, is 6. In this case, we can say k= 6 and hence, \(x^2\) = 3600, which leads us to x = 60. So, 60 is certainly a possible value of x. Now, 60 is the value of x as per statement I. Therefore, any option that does not contain statement I, can be eliminated. So, options B and C go out. We are left with options A, D and E.

Let’s look at statement II. As per statement II, x = 200, which means \(x^2\) = 40000. But, 40000 is not a multiple of 3 and hence cannot be a multiple of 600 (which is a multiple of 3). This violates the condition given in the question that \(x^2\) is divisible by 40 and 75. So, we can safely conclude that statement II cannot be true at all. We can now eliminate option E, we are left with options A and D.

If k = \(2^5\) * \(3^1\), \(x^2\) = 600k = 600 * 96 = 57600. Therefore, x = 240. Option A can now be eliminated.

The correct answer is option D.

All perfect squares greater than 1, will always have at least ONE pair of prime factors. This is the concept that needs to be applied in this question, to find out the value of k, because clearly, 600 does not contain pairs of all the prime factors that you see. Also, x having exactly 3 distinct prime factors is obvious from the fact that 600 has 3 distinct prime factors.

This problem should take you close to 2 minutes to solve, considering the fact that you will have to plug in different values for k.

Hope this helps!
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Re: x^2 is divisible by both 40 and 75. If x has exactly three distinct  [#permalink]

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New post 01 Jun 2019, 06:22
Bunuel wrote:
x is a positive integer, and \(x^2\) is divisible by both 40 and 75. If x has exactly three distinct prime factors, which of the following could be the value of x?

I. 60
II. 200
III. 240

A. I only
B. II only
C. III only
D. I and III only
E. I, II and III


40 = 2^3 x 5^1

75 = 5^2 x 3^1

So, the LCM of 40 and 75 is 2^3 x 3^1 x 5^2, and since x^2 is divisible by 2^3 x 3^1 x 5^2, we need to make the exponents of these prime factors (2, 3 and 5) even, so x^2 is at least 2^4 x 3^2 x 5^2.

Thus, the square root of 2^4 x 3^2 x 5^2 is 2^2 x 3 x 5 = 60. So we see that x can be any multiple of 60.

Answer: D
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Re: x^2 is divisible by both 40 and 75. If x has exactly three distinct   [#permalink] 01 Jun 2019, 06:22
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