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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 24 Apr 2008, 07:40
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

Answer: (B) -2.

Last edited by shobuj on 24 Apr 2008, 10:14, edited 1 time in total.
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Re: do it ? [#permalink] New post 24 Apr 2008, 09:29
Are you sure the question is complete the answer B fails for X=6
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Re: walker pls do it ? [#permalink] New post 24 Apr 2008, 10:26
shobuj 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

Answer: (B) -2.


I agree w/prasannar, something's missing from the question. Are you sure the question isn't something like:

X + (X - 1) + (X - k)?

That would yield B as the answer.
-------------------------------

Anyway, if it's the above terms, then just combine to get 3X - 1 - k, eliminate 3X since it's divisible by 3, plug and test with -1 - k to see what isn't divisible by 3.
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Re: walker pls do it ? [#permalink] New post 24 Apr 2008, 11:31
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

guys i have done a little bit about this:

if we put k=-1 we get:

X(x-1)(X+1) rearrange:(x-1)X(X+1)

so it looks like a sequenc,

if we assume that X =2 and put number from the answer then we get:
(x – 1)x(x – k)
k=5 =1.2.-3
k=2 =1.2.0
k=-1 =1.2.3
k=-4 =1.2.6
but when we put
k=-2 =1.2.4 not satisfied

but the stem says that x is and integer and 0 is an integer if we put 0 in this term than anything is divisable by 3, becaz 0 is divisable by 3

now i m also a little bit confused.

thanks
shobuj
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Re: walker pls do it ? [#permalink] New post 24 Apr 2008, 11:38
Where did you get the question from?
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Re: walker pls do it ? [#permalink] New post 24 Apr 2008, 11:45
i agree with B..

pick x=2..

1*2*4 isnt divisble by 3...

pick x=5

4*5*7..isnt divisble by 3...
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Re: walker pls do it ? [#permalink] New post 24 Apr 2008, 11:51
kiddrek i got it from another site and i think that walker can explain me much more better way

and the question is a valid one

but now can u clear me that in this question can we assume x = 0 or not ?

tanks
shobuj
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Re: walker pls do it ? [#permalink] New post 24 Apr 2008, 12:00
haha..ur pushing ur luck..yes for x=0..everything is divisble..

so poorly written question..but suppose it said x is a positive integer..then b would the right answer..

otherwise..Kidrecks stem is also valid..

shobuj wrote:
kiddrek i got it from another site and i think that walker can explain me much more better way

and the question is a valid one

but now can u clear me that in this question can we assume x = 0 or not ?

tanks
shobuj
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Re: walker pls do it ? [#permalink] New post 24 Apr 2008, 12:11
shobuj wrote:
kidderek i got it from another site and i think that walker can explain me much more better way

and the question is a valid one

but now can u clear me that in this question can we assume x = 0 or not ?

tanks
shobuj


As the questions reads, we can assume x = 0, negative, any whole number. But as prasannar has said, if x=3 or any number divisible by 3, then the entire product of x(x-1)(x-k) will be divisible by 3, irrespective of the value of k.

That is why I think my divisibility of SUM, X + (X-1) + (X-k), instead of product, makes more sense. It can't just be a coincidence that the answer choice fits too, can it?
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Re: walker pls do it ? [#permalink] New post 24 Apr 2008, 12:55
hmm
thanks

i pick it from manhattan GMAT forum?

thanks

i will send the link later
thanks
shobuj
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Re: walker pls do it ? [#permalink] New post 24 Apr 2008, 17:06
shobuj 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

Answer: (B) -2.



This is what I tried....

for term x(x-1)
pick x = 10 , expression becomes 10*9*(x-k) , this expression is anyway divisible by 3
so 10 is not a good pick to find out k

pick x = 9...again not good pick as 9 itself is divisible by 3

pick 8 , expression becomes...8*7*(x-k)...so (x-k) has to be multiple of 3.

If you try options now...all except B makes (x-k) multiple of 3.

So answer is B.
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Re: walker pls do it ? [#permalink] New post 24 Apr 2008, 21:54
yes

that what i stated before but my question lies on
what about when x =0

and for this question i think
they
want if x greater than 0 alll digit satisfy the sequence

bcause any threes term sequence is divisable by 3 equence
1,2.3
2,3,4
4,5,6
-1,0,1
-2,-1,0
-3,-2,-1
-1,1,3 diff=2

thanks
shobuj
Re: walker pls do it ?   [#permalink] 24 Apr 2008, 21:54
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