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If x is an integer, then x(x – 1)(x – k)

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by GMATGuruNY » Thu Aug 04, 2016 5:28 am
Needgmat wrote:If x is an integer, then x(x - 1)(x - k) must be evenly divisible by three when k is any of the following values EXCEPT

A) -4
B) -2
C) -1
D) 2
E) 5
We can PLUG IN THE ANSWERS, which represent the value of k.

Since the goal is show that (x)(x-1)(x-k) does NOT have to be divisible by 3, let x=5, with the result that x and x-1 will each not be divisible by 3:
(x)(x-1)(x-k) = (5)(5-1)(5-k) = (5)(4)(5-k).
When the correct answer choice is plugged in, (5)(4)(5-k) will NOT be divisible by 3.

A: k=-4
Here, (5)(4)(5-k) = (5)(4)(9).
The resulting product is a multiple of 3.

B: k=-2
Here, (5)(4)(5-k) = (5)(4)(7).
The resulting product is NOT a multiple of 3.

The correct answer is B.
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by Needgmat » Thu Aug 04, 2016 8:43 am
Since the goal is show that (x)(x-1)(x-k) does NOT have to be divisible by 3, let x=5, with the result that x and x-1 will each not be divisible by 3:
(x)(x-1)(x-k) = (5)(5-1)(5-k) = (5)(4)(5-k).
When the correct answer choice is plugged in, (5)(4)(5-k) will NOT be divisible by 3.
Hi GMATGuruNY ,

Thank you so much for your explanation.

Just a quick question, is there any particular reason to take x=5 or I can take any number, which is not divisible by 3?

Please explain.

Many thanks in advance.

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by [email protected] » Thu Aug 04, 2016 9:56 am
Hi Needgmat,

This question is based on the exact same concept as an OG question:

OG13/GMAT2015 PS #87
GMAT2016 PS #106
GMAT2017 PS #119

When dealing with 3 consecutive integers, exactly one of the 3 will be divisible by the number 3:

eg
1, 2, 3
2, 3, 4
3, 4, 5
4, 5, 6
5, 6, 7
Etc.

Thus, when multiplying 3 consecutive integers, we'll end up with a product that is divisible by 3 (since there will be at least one 3 in the prime-factorization of that product). In this prompt, we have (X) and (X-1), so we know that we have two consecutive integers. However, we don't know whether one of those two is the one that's divisible by 3 or if they're the two numbers that are not (and the third value is the one that's divisibly by 3.

IF... the third term was either (X+1) or (X-2), then we'd be guaranteed to have 3 consecutive integers...

(X+1)(X)(X-1)

or

(X)(X-1)(X-2)

Those are not the only options though. Any term that is "3 away" from the third term would ALSO be a multiple of 3.

eg.

1, 2, 3
1, 2, 6
1, 2, 9

6, 7, 8
3, 7, 8
9, 7, 8
Etc.

So instead of (X+1) and (X-2) we could have...

(X+4)
(X+7)
(X+10)
Etc.

(X-2)
(X-5)
(X-8)
Etc.

Knowing that pattern, it doesn't take much work to find the correct answer...

Final Answer: B

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by GMATGuruNY » Fri Aug 05, 2016 6:11 am
Needgmat wrote:Hi GMATGuruNY ,

Thank you so much for your explanation.

Just a quick question, is there any particular reason to take x=5 or I can take any number, which is not divisible by 3?
The goal is to show that (x)(x-1)(x-k) does NOT have to be divisible by 3.
To this end, we can test any integer for x such that neither x nor x-1 will be divisible by 3.
I tested x=5 because it is greater than or equal to all of the answer choices, with the result that 5-k will be nonnegative in every case.
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by Matt@VeritasPrep » Fri Aug 05, 2016 2:42 pm
Needgmat wrote: Just a quick question, is there any particular reason to take x=5 or I can take any number, which is not divisible by 3?
It might be easiest to think about remainders when dividing by 3. Integers fall into three groups when divided by 3:

Remainder 0: 0, 3, 6, 9, 12, ...
Remainder 1: 1, 4, 7, 10, 13, ...
Remainder 2: 2, 5, 8, 11, 14, ...

So if you're multiplying three numbers together, you know that one of them will be divisible by three if you have each possible remainder represented.

For instance, suppose I have an integer x, and I don't know its remainder by 3. I know that (x - 1) will have a different remainder from x, and I also know that (x - 4) will have the SAME remainder as (x - 1), because (x - 1) and (x - 4) differ by 3 and hence appear in the same remainder group. So if I do

x * (x - 4)

I know I have two remainders represented.

Now all I need is the third remainder (which could be (x - 2) or (x - 5) or (x - 8), etc.) and I'm set! I know I'll have SOME integer that divides by 3.
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