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Manager Joined: 07 Feb 2010
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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

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 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.

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]

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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]

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This is my 1st post finally thought of jumping in instead of just being an observer 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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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.

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]

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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]

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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]

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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.

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Re: If x is an integer, then x(x – 1)(x – k) must be evenly divisible by  [#permalink]

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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.

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...
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Re: If x is an integer, then x(x – 1)(x – k) must be evenly divisible by  [#permalink]

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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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Re: If x is an integer, then x(x – 1)(x – k) must be evenly divisible by  [#permalink]

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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!

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Re: If x is an integer, then x(x – 1)(x – k) must be evenly divisible by  [#permalink]

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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]

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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]

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Thanks EvaJager, I was always thought it is not good way to solve and I should learn the better way.
Thanks a lot... Director Joined: 22 Mar 2011
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Re: If x is an integer, then x(x – 1)(x – k) must be evenly divisible by  [#permalink]

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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?
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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]

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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]

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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]

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VeritasPrepKarishma wrote:
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.

Quote:
Plz Could you please explain how x-5 will also be a multiple of 3. I couldnot understand that part.

If (x - 2) is a multiple of 3, (x - 5), a number 3 places away from (x - 5) will also be divisible by 3.

Say (x - 2) = 9 (a multiple of 3)
then (x - 5) = 6 (previous multiple of 3)
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Re: If x is an integer, then x(x – 1)(x – k) must be evenly divisible by  [#permalink]

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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.

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]

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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.

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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