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Re: consecutive integers: product divisible by three [#permalink]
23 May 2010, 01:11

8

This post received KUDOS

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

_________________

press kudos, if you like the explanation, appreciate the effort or encourage people to respond.

Re: consecutive integers: product divisible by three [#permalink]
23 May 2010, 02:36

5

This post received KUDOS

2

This post was BOOKMARKED

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

Re: consecutive integers: product divisible by three [#permalink]
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???

Re: consecutive integers: product divisible by three [#permalink]
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 ..

_________________

press kudos, if you like the explanation, appreciate the effort or encourage people to respond.

Re: consecutive integers: product divisible by three [#permalink]
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.

Re: consecutive integers: product divisible by three [#permalink]
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.

Re: consecutive integers: product divisible by three [#permalink]
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.

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