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Re: walker pls do it ? [#permalink]
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.

Re: walker pls do it ? [#permalink]
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

Re: walker pls do it ? [#permalink]
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?