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M70-01

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M70-01  [#permalink]

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New post 03 Sep 2018, 01:50
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

56% (00:47) correct 44% (01:05) wrong based on 18 sessions

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Math Expert
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Re M70-01  [#permalink]

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New post 03 Sep 2018, 01:50
Official Solution:


If \(x\) is a positive integer, which of the following is not necessarily a divisor of \(x(x + 1)(x + 2)\)?


A. \(x\)
B. 2
C. 3
D. 4
E. 6


We’ll go for ALTERNATIVE since we’re asked about an expression with variables.

When looking at the answers, \(x\) is definitely a divisor, since it is a multiplier in the expression itself. But what about the rest of the answers? Since this is an expression with only letters, we can pick a number: \(x = 1\) gives us \(1 × 2 × 3 = 6\). So 2, 3, and 6 are all divisors of 6, but 4 isn’t.


Answer: D
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Re: M70-01  [#permalink]

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New post 15 Sep 2018, 21:19
I don't understand this question. If we are allowed to "pick a number" and I choose x=4 then the solution is divisible by all the answer choices. I must try another number that isn't divisible by all the answers?
x=1
1(1+1)(1+2)
(1)(2)(3)= 6 (which is not divisible by 4

but what if I choose x=4

4(4+1)(4+2)
4(5)(6)= 120 (divisible by 3,6,and 4.

Do I have to chose a number that is not a possible answer?
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Re: M70-01  [#permalink]

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New post 16 Sep 2018, 11:31
Correct. Keep in mind, the question reads: "which of the following is not NECESSARILY a divisor of x(x+1)(x+2)x(x+1)(x+2)?" That means there will be some answer choices where they all work. But you need to find the one where one of them doesn't work. Even though 4 is divisible when X=4, it is not divisible when X=1. The other answer choices will be divisible regardless of what X is equal to. So if you try X=4 and can't determine the correct answer, try incrementing your X value by one until you find the correct answer.
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Re: M70-01  [#permalink]

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New post 13 Oct 2018, 12:37
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Re: M70-01  [#permalink]

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New post 28 Nov 2018, 09:27
It will definitely be divisible by 2 because any 3 consecutive numbers would contain an even number (thus the product would be even); and any 3 consecutive numbers would contain a multiple of 3 (thus divisible by 3). Since it is an even number that can be divided by 3, it would definitely be divisible by 6. But 4 is not a definite divisor of the product of any 3 consecutive numbers.
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Re: M70-01   [#permalink] 28 Nov 2018, 09:27
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