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

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If x is an integer, then x(x - 1)(x - k) must be evenly [#permalink] New post 22 May 2010, 04:15
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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

Open discussion of this question is here: if-x-is-an-integer-then-x-x-1-x-k-must-be-evenly-divisible-126853.html
[Reveal] Spoiler: OA

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Last edited by Bunuel on 12 May 2012, 11:39, edited 1 time in total.
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Re: consecutive integers: product divisible by three [#permalink] New post 23 May 2010, 01:11
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i was hoping to get SOME replies on the post but anyway i'll share a very nice approach to handle such questions.

EXCEPT in the question tells us that all answer choices are alike other than one exception, the odd one out. If we can find out the odd one, we don't need to do any calculation and we can still answer this under 15secs.

now the question talks about dividing the product by 3 ... so 4/5 answers choices should be related to 3 .... let's see the difference between the answer choices

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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Last edited by dimitri92 on 24 May 2010, 06:17, edited 2 times in total.
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Re: consecutive integers: product divisible by three [#permalink] New post 23 May 2010, 02:36
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Although you method is shortest, i would prefer surest way...
If x or x-1 is divisible by 3 ==> x(x - 1)(x - k) is divisible by 3
Assign any arbitrary number to x such that x and x-1 are not disible by 3
say x=5, x-1 = 4 ==> 20*(5-k)
assign answer choices to k..
for k=-4, ==> 20*(9) ==> divisible by 3
for k=-2 ==> 20(7) ==> NOT divisible by 3

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Re: consecutive integers: product divisible by three [#permalink] New post 23 May 2010, 02:20
Hi dimitri,
Amazing approach. I think this is one of the best approches i have ever encountered. +1 kudo for you...
But one question for you... what if one of the choice is not following the pattern?? and still that may not be the answer???
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Re: consecutive integers: product divisible by three [#permalink] New post 23 May 2010, 12:27
here's another approach

x(x - 1)(x - k)
all three are consecutive, so the product MUST be a multiple of 3

we don't know the value of k just yet ... so let's extend the series ... the extension itself reveals the answers

..(x-5)..(x-2)
(x-1)x(x+1)..(x+4)..

we can see the possible values of k too from the series

k = 2 OR 2+3n [2 & 5]
k = -1 OR -1+3n [-1 & -4]

B i.e. -2 does not fit in any value of k

so B it is

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Re: consecutive integers: product divisible by three [#permalink] New post 24 May 2010, 04:14
dimitri 92

could you explain how do you come up with expansion series and how do you pick those for K? I'm still confused on your method

thank you very much.
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Re: consecutive integers: product divisible by three [#permalink] New post 24 May 2010, 06:14
abcd1 wrote:
dimitri 92

could you explain how do you come up with expansion series and how do you pick those for K? I'm still confused on your method

thank you very much.


the given series: x(x - 1)(x - k)
but we don't know the value of k ... for (x-k) to be a multiple of 3, it must be consecutive to x(x-1) .. so it must be either (x-2) or (x+1) i.e. value of k can be 2 or -1.

there are two ways of expanding the series now
..(x-8)..(x-5)..(x-2)
(x-1)x..
OR
..
(x-1)x(x+1)..(x+4)..(x+7)..

from the above expansion you can see that the following are possible values of k
2, 5, 8, 11 ..
-1, -4, -7 ..


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Re: consecutive integers: product divisible by three [#permalink] New post 24 May 2010, 18:16
thank you very much ^_^ dimitri 92
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Consecutive integers [#permalink] New post 15 Jun 2010, 23:54
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
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Re: Consecutive integers [#permalink] New post 16 Jun 2010, 00:01
The easiest way is to plug in an integer which depends on (x-k) for the expression to be divisible by 3.

Example, pick 5:
The expression becomes: 5* 4 * (5 - k)

Now from the 5 values of k, see which value does not give you a multiple of 3.

A. 5 -(-4) = 9
B. 5 -(-2) = 7 ... Bingo, the expression would be equal to 140 which is not divisible by 3. On test day - stop.
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Re: consecutive integers: product divisible by three [#permalink] New post 20 Jun 2010, 23:08
If your mind blanked out during the test, the easiest thing to do is to substitute each option into K. Of course, you must understand the concept that for a consecutive of three, x+1 occupies the same position as x+(1+6), x+(1+3), x+(1-3) or x+(1-6) and so on. All choices except -2 will give the consecutive numbers. Substituting -2 will give x(x-1)(x--2) = x(x-1)(x+2) which is not a three consecutive numbers.
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Re: consecutive integers: product divisible by three [#permalink] New post 12 May 2012, 11:00
the given series: x(x - 1)(x - k)
but we don't know the value of k ... for (x-k) to be a multiple of 3, it must be consecutive to x(x-1) .. so it must be either (x-2) or (x+1) i.e. value of k can be 2 or -1.


Dimitri, for (x-k) to be a multiple of 3, why MUST it be CONSECUTIVE to x(x-1)?

example: if x=5 5(5-1)(5-20)= 300 which is divisible by 3 even tho 5x4x15 are not consecutive integers.

I realize that assuming they are consecutive integers helps you solve the problem but I just wanted to be clear that (x-k) can be a multiple of 3 without it being consecutive to x(x-1).

Please let me know if i'm missing something, thanks.
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Re: consecutive integers: product divisible by three [#permalink] New post 12 May 2012, 11:38
Expert's post
ctiger100 wrote:
the given series: x(x - 1)(x - k)
but we don't know the value of k ... for (x-k) to be a multiple of 3, it must be consecutive to x(x-1) .. so it must be either (x-2) or (x+1) i.e. value of k can be 2 or -1.


Dimitri, for (x-k) to be a multiple of 3, why MUST it be CONSECUTIVE to x(x-1)?

example: if x=5 5(5-1)(5-20)= 300 which is divisible by 3 even tho 5x4x15 are not consecutive integers.

I realize that assuming they are consecutive integers helps you solve the problem but I just wanted to be clear that (x-k) can be a multiple of 3 without it being consecutive to x(x-1).

Please let me know if i'm missing something, thanks.


You are right x, (x – 1), and (x – k) are not necessarily consecutive integers.

Open discussion of this question is here: if-x-is-an-integer-then-x-x-1-x-k-must-be-evenly-divisible-126853.html
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Re: consecutive integers: product divisible by three   [#permalink] 12 May 2012, 11:38
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