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If x is an integer, then x(x – 1)(x – k) must be evenly divisible by

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anilnandyala
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If x is an integer, then x(x – 1)(x – k) must be evenly divisible by [#permalink] New post   Updated on: 23 Sep 2019, 03:50
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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
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B

Originally posted by anilnandyala on 15 Dec 2010, 07:19.
Last edited by Bunuel on 23 Sep 2019, 03:50, edited 2 times in total.
Renamed the topic and edited the question.
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If x is an integer, then x(x – 1)(x – k) must be evenly divisible by [#permalink] New post   31 Jan 2012, 16:06
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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 have the product of 3 integers: (x-1)x(x-k).

Note that the product of 3 integers is divisible by 3 if at least one multiple is divisible by 3. Now, to guarantee that at least one integer out of x, (x – 1), and (x – k) is divisible by 3 these numbers must have different remainders upon division by 3, meaning that one of them should have remainder of 1, another reminder of 2 and the last one remainder of 0, so be divisible by 3.

Next, if k=-2 then we'll have (x-1)x(x+2)=(x-1)x(x-1+3) --> which means that (x-1) and (x+2) will have the same remainder upon division by 3. Thus for k=-2 we won't be sure whether (x-1)x(x-k) is divisible by 3.

Answer: B.

30 second approach: 4 out of 5 values of k from answer choices must guarantee divisibility of some expression by 3. Now, these 4 values of k in answer choices must have some pattern: if we get rid of -2 then -4, -1, 2, and 5 creating arithmetic progression with common difference of 3, so -2 is clearly doesn't belong to this pattern.

Hope it helps.
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Re: If x is an integer, then x(x – 1)(x – k) must be evenly divisible by [#permalink] New post   04 Jan 2011, 19:49
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anilnandyala 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


-4
-2
-1
2
5


I am providing the theoretical explanation below. Once you get it, you can solve such questions in a few seconds in future!

Notice a few things about integers:
-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16......

Every number is a multiple of 1
Every second number is a multiple of 2
Every third number is a multiple of 3
Every fourth number is a multiple of 4 and so on...

So if I pick any 3 consecutive integers, one and only one of them will be a multiple of 3: e.g. I pick 4, 5, 6 (6 is a multiple of 3) or I pick 11, 12, 13 (12 is a multiple of 3) etc..

x(x - 1)(x - k) will be evenly divisible by 3 if at least one of x, x - 1 and x - k is a multiple of 3. We know from above, (x - 2)(x - 1)x will have a multiple of 3 in it. Also, (x-1)x(x + 1) will have a multiple of 3 in it because they both are products of 3 consecutive integers. So k can be 2 or -1. Eliminate these options.
Now let me write down consecutive integers around x:

(x-5), (x - 4), (x - 3), (x - 2), (x - 1), x, (x + 1), (x + 2), (x + 3), (x + 4), (x + 5) etc

(x - 2)(x - 1)x will have a multiple of 3 in it. x could be the multiple of 3, (x - 1) could be the multiple of 3 or (x - 2) could be the multiple of 3, in which case (x - 5) will also be a multiple of 3.
So in any case, (x - 5)(x - 1)x will have a multiple of 3 in it. So k can be 5.

Similarly, (x-1)x(x + 1) will have a multiple of 3 in it. x could be the multiple of 3, (x - 1) could be the multiple of 3 or (x + 1) could be the multiple of 3, in which case (x + 4) will also be a multiple of 3.
So in any case, (x - 1)x(x + 4) will have a multiple of 3 in it. So k can be -4.

We cannot say whether (x-1)x(x + 2) will have a multiple of 3 in it and hence if k = -2, we cannot say whether the product is evenly divisible by 3.

Answer (B).
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Re: If x is an integer, then x(x – 1)(x – k) must be evenly divisible by [#permalink] New post   31 Jan 2012, 16:17
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Hi, there. I'm happy to help with this. :)

The rule that the product of three consecutive integers is a good start, but not the be all and end all.

Think about it this way:

number = x(x – 1)(x – k)

So far, we have integer x and one less that it (x - 1), so we could go down one more, or up one from x --

k = 2 ----> x(x – 1)(x – 2)

k = -1 ----> x(x – 1)(x + 1)

Now, we don't know which of the three factors are divisible by 3 -- x, or (x - 1), or the (x - k). If it's either of the first two, then we're golden, and k doesn't matter. But pretend that neither x nor (x - 1) is divisible by 3, then we are dependent on that last factor. Well, if (x - 2) is a multiple of three, we should be able to add or subtract three and still get a multiple of three.

(x - 2) - 3 = (x - 5)

(x - 5) - 3 = (x - 8)

(x - 2) + 3 = (x + 1), which we have already

(x + 1) + 3 = (x + 4)

(x + 3) + 3 = (x + 7)

So, for divisibility purposes, (x - 8), (x - 5), (x - 2), (x + 1), (x + 4), (x + 7) are all equivalent -- if any one of them is a multiple of three, all the others are. (You can check that the difference between any two is a multiple of 3.) BTW, if they are not divisible by three, then they all would have equal remainders if divided by three.

Back to the question:
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

All of those choices give us a term on our list except for (B) -2.

BTW, notice all the answer choices are spaced apart by three except for (B).

Does that make sense? Please do not hesitate to ask if you have any questions.

Mike :)
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If x is an integer, then x(x – 1)(x – k) must be evenly divisible by [#permalink] New post   15 Dec 2010, 07:52
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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 have the product of 3 integers: (x-1)x(x-k).

Note that the product of 3 integers is divisible by 3 if at least one multiple is divisible by 3. Now, to guarantee that at least one integer out of x, (x – 1), and (x – k) is divisible by 3 these numbers must have different remainders upon division by 3, meaning that one of them should have remainder of 1, another reminder of 2 and the last one remainder of 0, so be divisible by 3.

Next, if k=-2 then we'll have (x-1)x(x+2)=(x-1)x(x-1+3) --> which means that (x-1) and (x+2) will have the same remainder upon division by 3. Thus for k=-2 we won't be sure whether (x-1)x(x-k) is divisible by 3.

Answer: B.

30 second approach: 4 out of 5 values of k from answer choices must guarantee divisibility of some expression by 3. Now, these 4 values of k in answer choices must have some pattern: if we get rid of -2 then -4, -1, 2, and 5 creating arithmetic progression with common difference of 3, so -2 is clearly doesn't belong to this pattern.

Hope it helps.
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Re: If x is an integer, then x(x – 1)(x – k) must be evenly divisible by [#permalink] New post   31 Jan 2012, 16:34
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This is my 1st post :) finally thought of jumping in instead of just being an observer :-D
I attacked this problem in a simple way. As it states it is divisible by 3
that means both x & (x-1) cannot be a multiple of 3 otherwise whatever the value of k it will be still divisible by 3
so plugging in number i chose 5 in this case you can establish answer is -2 does not fit...
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Re: If x is an integer, then x(x – 1)(x – k) must be evenly divisible by [#permalink] New post   03 Jan 2011, 22:00
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Hi bunuel,
I don't understand the problem language, it says

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

how does it matter whats the value of K, i can choose x = 3 and the expression will always be divisible by 3.

Am i missing any minor yet important point?
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Re: If x is an integer, then x(x – 1)(x – k) must be evenly divisible by [#permalink] New post   04 Jan 2011, 03:22
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vjsharma25 wrote:
Hi bunuel,
I don't understand the problem language, it says

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

how does it matter whats the value of K, i can choose x = 3 and the expression will always be divisible by 3.

Am i missing any minor yet important point?


Stem says: "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"

The important word in the stem is "MUST", which means that we should guarantee the divisibility by 3 no matter the value of x (for ANY integer value of x), so you cannot arbitrary pick its value.

Hope it's clear.
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Re: If x is an integer, then x(x – 1)(x – k) must be evenly divisible by [#permalink] New post   07 Jan 2011, 23:49
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Bunuel wrote:
anilnandyala 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


-4
-2
-1
2
5


We have the product of 3 integers: (x-1)x(x-k).

Note that the product of 3 integers is divisible by 3 if at least one multiple is divisible by 3. Now, to guarantee that at least one integer out of x, (x – 1), and (x – k) is divisible by 3 these numbers must have different remainders upon division by 3, meaning that one of them should have remainder of 1, another reminder of 2 and the last one remainder of 0, so be divisible by 3.

Next, if k=-2 then we'll have (x-1)x(x+2)=(x-1)x(x-1+3) --> which means that (x-1) and (x+2) will have the same remainder upon division by 3. Thus for k=-2 we won't be sure whether (x-1)x(x-k) is divisible by 3.

Answer: B.

30 second approach: 4 out of 5 values of k from answer choices must guarantee divisibility of some expression by 3. Now, these 4 values of k in answer choices must have some pattern: if we get rid of -2 then -4, -1, 2, and 5 creating arithmetic progression with common difference of 3, so -2 is clearly doesn't belong to this pattern.

Hope it helps.



Bunnel,
The second approach is too good...
Very helpful,...
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Re: If x is an integer, then x(x – 1)(x – k) must be evenly divisible by [#permalink] New post   04 Oct 2012, 23:10
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enigma123 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

The OA is B. I am trying to use the concept of consecutive numbers but got stuck. Can someone please help?

Question says x(x – 1)(x – k) must be evenly divisible by three which means x(x-1) (x-k) should be consecutive.


Since this is a multiple choice GMAT question, you can pick a particular value for x such that neither x, nor x-1 is divisible by 3 and start checking the answers.
In the given situation, choose for example x = 2 and check when 2 - k is not divisible by 3.
(A) 2 - (-4) = 6 NO
(B) 2 - (-2) = 4 BINGO!

Answer B.
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Re: If x is an integer, then x(x – 1)(x – k) must be evenly divisible by [#permalink] New post   05 Oct 2012, 03:37
I too solved it using a value for x =2, but I am not sure if is it better to solve using value for such questions or otherwise.
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Re: If x is an integer, then x(x – 1)(x – k) must be evenly divisible by [#permalink] New post   05 Oct 2012, 03:56
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Avantika5 wrote:
I too solved it using a value for x =2, but I am not sure if is it better to solve using value for such questions or otherwise.


On the GMAT, is definitely the fastest way to solve it. Being a multiple choice question, you can be sure that there is a unique correct answer. And in the given situation, the only issue is to choose for x values such that neither x, nor x - 1 is divisible by 3.
It won't harm to understand and know to use the properties of consecutive integers presented in the other posts . They can be useful any time.
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Re: If x is an integer, then x(x – 1)(x – k) must be evenly divisible by [#permalink] New post   06 Oct 2012, 01:10
Thanks EvaJager, I was always thought it is not good way to solve and I should learn the better way.
Thanks a lot... :)
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Re: If x is an integer, then x(x – 1)(x – k) must be evenly divisible by [#permalink] New post   Updated on: 06 Oct 2012, 02:46
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Avantika5 wrote:
Thanks EvaJager, I was always thought it is not good way to solve and I should learn the better way.
Thanks a lot... :)


Better is a relative word...Mathematicians always try to prove and justify everything in a formal, logical way.
But GMAT is not testing mathematical abilities per se. If they wanted so, the questions would have been open and not multiple choice.
Have a flexible mind, think out of the box. GMAT is not a contest for the most beautiful, elegant, mathematical solution...
Get the correct answer as quickly as possible, and go to the next question without any feeling of guilt...:O)

Though, as I said, try to understand the other properties of the integer numbers, they can become handy and also, because they are so beautiful! Isn't Mathematics wonderful?

Originally posted by EvaJager on 06 Oct 2012, 02:10.
Last edited by EvaJager on 06 Oct 2012, 02:46, edited 1 time in total.
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Re: If x is an integer, then x(x – 1)(x – k) must be evenly divisible by [#permalink] New post   08 Mar 2014, 05:51
Hats off to Bunuel for the 30 sec. Approach!Couldn't visualize that solution!

Posted from my mobile device
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Re: If x is an integer, then x(x – 1)(x – k) must be evenly divisible by [#permalink] New post   02 Aug 2014, 01:42
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anilnandyala 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


Since x(x-1)(x-k) is divisible by 3, take a case when x(x-1) is not divisible by 3 and so (x-k) has to be divisible by 3.
Let us take x=8 and x-1=7. Only for the second option we do not get x-k divisible by 3.
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Re: If x is an integer, then x(x – 1)(x – k) must be evenly divisible by [#permalink] New post   03 Oct 2014, 08:45
Bunuel wrote:
anilnandyala 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


-4
-2
-1
2
5


We have the product of 3 integers: (x-1)x(x-k).

Note that the product of 3 integers is divisible by 3 if at least one multiple is divisible by 3. Now, to guarantee that at least one integer out of x, (x – 1), and (x – k) is divisible by 3 these numbers must have different remainders upon division by 3, meaning that one of them should have remainder of 1, another reminder of 2 and the last one remainder of 0, so be divisible by 3.

Next, if k=-2 then we'll have (x-1)x(x+2)=(x-1)x(x-1+3) --> which means that (x-1) and (x+2) will have the same remainder upon division by 3. Thus for k=-2 we won't be sure whether (x-1)x(x-k) is divisible by 3.

Answer: B.

30 second approach: 4 out of 5 values of k from answer choices must guarantee divisibility of some expression by 3. Now, these 4 values of k in answer choices must have some pattern: if we get rid of -2 then -4, -1, 2, and 5 creating arithmetic progression with common difference of 3, so -2 is clearly doesn't belong to this pattern.

Hope it helps.



Hi Bunuel,
does "evenly divisible" mean that the dividend on being divided by 3, leave a quotient that is even??
please correct me if i am wrong.
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Re: If x is an integer, then x(x – 1)(x – k) must be evenly divisible by [#permalink] New post   03 Oct 2014, 08:49
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arnabs wrote:
Bunuel wrote:
anilnandyala 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


-4
-2
-1
2
5


We have the product of 3 integers: (x-1)x(x-k).

Note that the product of 3 integers is divisible by 3 if at least one multiple is divisible by 3. Now, to guarantee that at least one integer out of x, (x – 1), and (x – k) is divisible by 3 these numbers must have different remainders upon division by 3, meaning that one of them should have remainder of 1, another reminder of 2 and the last one remainder of 0, so be divisible by 3.

Next, if k=-2 then we'll have (x-1)x(x+2)=(x-1)x(x-1+3) --> which means that (x-1) and (x+2) will have the same remainder upon division by 3. Thus for k=-2 we won't be sure whether (x-1)x(x-k) is divisible by 3.

Answer: B.

30 second approach: 4 out of 5 values of k from answer choices must guarantee divisibility of some expression by 3. Now, these 4 values of k in answer choices must have some pattern: if we get rid of -2 then -4, -1, 2, and 5 creating arithmetic progression with common difference of 3, so -2 is clearly doesn't belong to this pattern.

Hope it helps.



Hi Bunuel,
does "evenly divisible" mean that the dividend on being divided by 3, leave a quotient that is even??
please correct me if i am wrong.


No, evenly divisible means divisible without remainder, so simply divisible.
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Re: If x is an integer, then x(x – 1)(x – k) must be evenly divisible by [#permalink] New post   27 Oct 2014, 06:45
I tried the problem in a different way:

for any number to be divisible by 3, the sum of the integers in the number should be a factor of 3.

Taking the sum of x, (x-1) and (x-k) we have:

x+x-1+x-k = 3x-1-k

now looking at the choices

k = -4 => sum = 3x+3 --> divisible by 3
k = -2 => sum = 3x+1 --> not divisible by 3
k = -1 => sum = 3x --> divisible by 3
k = 2 => sum = 3x-3 --> divisible by 3
k = 5 ==> sum = 3x-6 --> divisible by 3

so answer choice is (b)
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Re: If x is an integer, then x(x – 1)(x – k) must be evenly divisible by [#permalink] New post   17 Jun 2015, 05:49
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santorasantu wrote:
Quote:
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



I tried the problem in a different way:

for any number to be divisible by 3, the sum of the integers in the number should be a factor of 3.

Taking the sum of x, (x-1) and (x-k) we have:

x+x-1+x-k = 3x-1-k

now looking at the choices

k = -4 => sum = 3x+3 --> divisible by 3
k = -2 => sum = 3x+1 --> not divisible by 3
k = -1 => sum = 3x --> divisible by 3
k = 2 => sum = 3x-3 --> divisible by 3
k = 5 ==> sum = 3x-6 --> divisible by 3

so answer choice is (b)


Just refining the highlighted language and presenting another view to del with this problem

CONCEPT1:For any number to be divisible by 3, the sum of the Digits of the Number should be a Multiple of 3.

CONCEPT2: Product of any three consecutive Integers always include one muliple of 3 hence product of any three consecutive Integers is always a Multiple of 3

CONCEPT3: If a Number "w" is a Multiple of 3 then any number at a difference of 3 or multiple of 3 from "w" will also be a multiple of 3 i.e. If w is a multiple of 3 then (w+3), (w-3), (w+6), (w-6) etc. will all be Multiples of 3

Here, I see that x(x – 1) is a product of two consecutive Integers but if another Number next to them is obtained then x(x – 1)(x – k) will certainly be a multiple of 3 [as per Concept2 mentioned above]

for x(x – 1)(x – k) to be a product of 3 consecutive integers,

(x – k) should be either (x - 2) i.e. k=2
OR
(x – k) should be either (x - 5) i.e. k=5 [Using Concept3]
OR
(x – k) should be either (x + 1) i.e. k=-1
OR
(x – k) should be either (x + 4) i.e. k=-4 [Using Concept3]

This Eliminates options A, C D and E

Answer: option
Show Spoiler
B
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Re: If x is an integer, then x(x – 1)(x – k) must be evenly divisible by [#permalink]
17 Jun 2015, 05:49
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