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y = x(x + 1)(x + 2) where x is an even integer. Which of the following

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Joined: 02 Sep 2009
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y = x(x + 1)(x + 2) where x is an even integer. Which of the following  [#permalink]

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08 Jul 2017, 05:34
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$$y = x(x + 1)(x + 2)$$ where x is an even integer. Which of the following must be true?

I. y is divisible by 6

II. y is divisible by 12

III. y is divisible by 18

A. I only
B. I and II only
C. I and III only
D. I, II, and III
E. None of the above

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Re: y = x(x + 1)(x + 2) where x is an even integer. Which of the following  [#permalink]

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08 Jul 2017, 05:41
Bunuel wrote:
$$y = x(x + 1)(x + 2)$$ where x is an even integer. Which of the following must be true?

I. y is divisible by 6

II. y is divisible by 12

III. y is divisible by 18

A. I only
B. I and II only
C. I and III only
D. I, II, and III
E. None of the above

Hi
Numbers are even * odd * even. So following properties
1) one even will be multiple of 2 and one of 4... ATLEAST
2) ATLEAST one will be multiple of 3..
3) so the product will be multiple of ATLEAST 2*4*3=24

Now 6 and 12 are factors of 24, so I and II MUST be true.
18 is NOT a factor of 24... So not a MUST

B
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Re: y = x(x + 1)(x + 2) where x is an even integer. Which of the following  [#permalink]

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08 Jul 2017, 05:50
1
chetan2u wrote:
Bunuel wrote:
$$y = x(x + 1)(x + 2)$$ where x is an even integer. Which of the following must be true?

I. y is divisible by 6

II. y is divisible by 12

III. y is divisible by 18

A. I only
B. I and II only
C. I and III only
D. I, II, and III
E. None of the above

Hi
Numbers are even * odd * even. So following properties
1) one even will be multiple of 2 and one of 4... ATLEAST
2) ATLEAST one will be multiple of 3..
3) so the product will be multiple of ATLEAST 2*4*3=24

Now 6 and 12 are factors of 24, so I and II MUST be true.
18 is NOT a factor of 24... So not a MUST

B

Although it does not make any difference, just want to highlight that since it is mentioned that x is an even integer which includes 0 as well and the value of y can be 0 which is considered divisible by all integers but none of the integers are considered divisible by "0"
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Re: y = x(x + 1)(x + 2) where x is an even integer. Which of the following  [#permalink]

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08 Jul 2017, 05:54
Bunuel wrote:
$$y = x(x + 1)(x + 2)$$ where x is an even integer. Which of the following must be true?

I. y is divisible by 6

II. y is divisible by 12

III. y is divisible by 18

A. I only
B. I and II only
C. I and III only
D. I, II, and III
E. None of the above

Let's plug in some numbers and check -

Let x = 2 ; So, $$y = 2*3*4$$ ( Divisible by 6 & 12 )
Let x = 4 ; So, $$y = 4*5*6$$ ( Divisible by 6 & 12 )
Let x = 6 ; So, $$y = 6*7*8$$ ( Divisible by 6 & 12 )

Thus, in each case y is divisible by 6 & 12, so the correct answer must be (B)
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y = x(x + 1)(x + 2) where x is an even integer. Which of the following  [#permalink]

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08 Jul 2017, 06:19
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Bunuel wrote:
$$y = x(x + 1)(x + 2)$$ where x is an even integer. Which of the following must be true?

I. y is divisible by 6

II. y is divisible by 12

III. y is divisible by 18

A. I only
B. I and II only
C. I and III only
D. I, II, and III
E. None of the above

$$y = x(x + 1)(x + 2)$$

$$y$$ is product of $$3$$ consecutive integers. Hence $$y$$ must be divisible by $$6$$.

Given $$x$$ is even integer. Hence $$y$$ must be divisible by $$12$$.

Let $$x = 2$$

$$y = 2(2+1)(2+2) = 2*3*4 = 24$$

$$24$$ is divisible by $$6$$ and $$12$$

$$24$$ is not divisible by $$18$$

I and II only . Answer (B)...
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Re: y = x(x + 1)(x + 2) where x is an even integer. Which of the following  [#permalink]

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29 Mar 2018, 08:43
0 is even and an integer too. IF x=0 option E becomes the answer. please clear the doubtBunuel
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Re: y = x(x + 1)(x + 2) where x is an even integer. Which of the following  [#permalink]

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29 Mar 2018, 08:48
shahul. wrote:
0 is even and an integer too. IF x=0 option E becomes the answer. please clear the doubtBunuel

0 is divisible by every integer except 0 itself: 0/integer = 0 = integer.
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Re: y = x(x + 1)(x + 2) where x is an even integer. Which of the following  [#permalink]

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29 Mar 2018, 09:00
Thanks bunuel , but if x =0, y will become zero and option d should be the answer. But b is the OA. Why is that ? Bunuel
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y = x(x + 1)(x + 2) where x is an even integer. Which of the following  [#permalink]

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29 Mar 2018, 09:04
1
shahul. wrote:
Thanks bunuel , but if x =0, y will become zero and option d should be the answer. But b is the OA. Why is that ? Bunuel

Notice that the question asks "which of the following statements must be true", not "which of the following statements could be true". While y could be divisible by any integer, it must be divisible only by some of the numbers.
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Re: y = x(x + 1)(x + 2) where x is an even integer. Which of the following  [#permalink]

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29 Mar 2018, 09:45
Bunuel wrote:
$$y = x(x + 1)(x + 2)$$ where x is an even integer. Which of the following must be true?

I. y is divisible by 6

II. y is divisible by 12

III. y is divisible by 18

A. I only
B. I and II only
C. I and III only
D. I, II, and III
E. None of the above

The only reason I marked (E) is that x can be "0". Point noted, 0 is divisible by every number except "0".
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Re: y = x(x + 1)(x + 2) where x is an even integer. Which of the following  [#permalink]

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01 Apr 2018, 23:05
Bunuel wrote:
$$y = x(x + 1)(x + 2)$$ where x is an even integer. Which of the following must be true?

I. y is divisible by 6

II. y is divisible by 12

III. y is divisible by 18

A. I only
B. I and II only
C. I and III only
D. I, II, and III
E. None of the above

IMOB By plug in numbers we know these are 3 consecutive integers where middle no. is odd and 1st and last no. is even so take any example 2,3,4 or 4,5,6 or 14,15,16 we can say only I and II situation is true.
Re: y = x(x + 1)(x + 2) where x is an even integer. Which of the following   [#permalink] 01 Apr 2018, 23:05
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