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

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If x is an integer, then x(x – 1)(x – k) must be evenly divi [#permalink] New post 15 Dec 2010, 06:19
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 09 Jul 2013, 08:59, edited 1 time in total.
Renamed the topic and edited the question.
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Re: x(x – 1)(x – k) [#permalink] New post 15 Dec 2010, 06:41
There is probably an easier way, but I just used the picking numbers option for this.

I chose x=2
2(1)(2-k) then just plugged in the answer choices for K until one wasn't evenly divisible by 3.

B gives you 8. 8/3 is not an integer.

B is the answer.
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Re: x(x – 1)(x – k) [#permalink] New post 15 Dec 2010, 06:52
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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


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: x(x – 1)(x – k) [#permalink] New post 03 Jan 2011, 21:00
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: x(x – 1)(x – k) [#permalink] New post 04 Jan 2011, 02:22
Expert's post
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 can not arbitrary pick its value.

Hope it's clear.
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Re: x(x – 1)(x – k) [#permalink] New post 04 Jan 2011, 07:46
OK. Now i get it.

Thanks Bunuel.
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Re: x(x – 1)(x – k) [#permalink] New post 04 Jan 2011, 08:41
To be divisible by 3, one of these sequences must be divisible by 3.

X(X-1) (X-k)

Any 3 sequence number will always be divisible by 3. So X(X-1) (x-2) is divisible by 3.

K = 2, divisible by 3
K= 5, also a sequence ( parallel ) divisible by 3
K= -1, sequence is (X-1) X (X+1) so divisible by 3
K= -4, also a sequence ( parallel ) divisible by 3
K=-2, not a sequence, may not be divisible by 3

So Answer is (B)
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Re: x(x – 1)(x – k) [#permalink] New post 04 Jan 2011, 18:49
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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: x(x – 1)(x – k) [#permalink] New post 05 Jan 2011, 00:29
a. -4
b. -2 [2 more than A]
c. -1 [3 more than A]
d. 2 [6 more than A]
e. 5 [9 more than A]

nice, so we do have a pattern ... 4 answers have a difference of a multiple of 3 except B ... 3, 6, 9 are all multiples of 3

so we can select B without solving much

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Re: x(x – 1)(x – k) [#permalink] New post 07 Jan 2011, 22:49
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.



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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 divi [#permalink] New post 29 Nov 2013, 20:22
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Re: If x is an integer, then x(x – 1)(x – k) must be evenly divi [#permalink] New post 07 Mar 2014, 03:52
Plugged in value of x = 8

It comes up 8 x 7 x (8-k)

Checking for each option available

-4 >>> 8+4 = 12.. Divisible by 3

-2 >>> 8+2 = 10.. Not divisible by 3

-1 >>> 8+1 = 9 .. Divisible by 3

2 >>> 8-2 = 6 .. Divisible by 3

5 >>> 8-5 = 3 .. Divisible by 3

Answer = B
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Re: If x is an integer, then x(x – 1)(x – k) must be evenly divi [#permalink] New post 08 Mar 2014, 04:51
Hats off to Bunuel for the 30 sec. Approach!Couldn't visualize that solution!

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Re: If x is an integer, then x(x – 1)(x – k) must be evenly divi [#permalink] New post 02 Aug 2014, 00:42
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 divi [#permalink] New post 01 Sep 2014, 23:21
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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.

Answer (B).

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 divi [#permalink] New post 03 Oct 2014, 07:45
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 divi [#permalink] New post 03 Oct 2014, 07:49
Expert's post
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 divi [#permalink] New post 03 Oct 2014, 07:56
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.



I am sorry but i did not really get the solution. A little more elaboration would help Bunuel. My main concern here is, if x(x-1)(x-k) were to be evenly divisible, then plugging any value for x(lets say 3) should make it evenly divisible by 3.
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Re: If x is an integer, then x(x – 1)(x – k) must be evenly divi [#permalink] New post 03 Oct 2014, 08:59
Expert's post
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.



I am sorry but i did not really get the solution. A little more elaboration would help Bunuel. My main concern here is, if x(x-1)(x-k) were to be evenly divisible, then plugging any value for x(lets say 3) should make it evenly divisible by 3.


Have you checked this: if-x-is-an-integer-then-x-x-1-x-k-must-be-evenly-divi-106310.html#p846137 ?
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Re: If x is an integer, then x(x – 1)(x – k) must be evenly divi [#permalink] New post 03 Oct 2014, 10:17
Bunuel wrote:
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 can not arbitrary pick its value.

Hope it's clear.



that was so helpful bunuel ,thank you so much!!! :-D
Re: If x is an integer, then x(x – 1)(x – k) must be evenly divi   [#permalink] 03 Oct 2014, 10:17
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