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Re: DS divisible by 6? [#permalink]
07 Feb 2005, 14:43

DLMD wrote:

N=k*(k+1)*(k-1), is n divisible by 6? 1) k is an positive integer. 2) k-1, k, and k+1 are consecutive integer

Ok, k is a positive integer in (1), so N would be 0*1*2, 1*2*3, ... etc. We know that product of any three consecutive positive integers is divisible by 6.

For (2), N could also be negative numbers, but still divisible by 6.

We can re-arrange the product to give: N=(k-1)(K)(K+1)

1) K is a positive integer.

Not sufficient.
If k=1, N=0*1*2=0 --> False
If K=2, N=1*2*3=6 --> True
If K=3, N=2*3*4=24 --> True
If K=5, N=4*5*6=120 --> True

The statement cannot be true as long as one integer does not answer our question.

2) K-1,K,K+1 are consecutive.
Not sufficient.

K can be 0,1,-1 which will result in 0 (not divisible by 6)
K can be any positive or negative integer (apart from 1,-1) and N will be divisible by 6

Combining both, we already see K can be 1 (divisible) or not 1 (2,3,4 etc, all of which gives a N which is divisible by 6)

Ans: E

** Note, the question doesn't states that K is an integer, so we can't assme K-1*K*K+1 will result in a even product which has a prime factor of 3. It is only the statements that suggest K is an integer

Honghu, to be honest, I'm not sure if 0 is divisible by any number ? I'm running on the logic that to be divisible you need to be able to factor it into prime numbers.
The number 0 seems to be sort of a special case. Let me check up my maths text and I'll let you know.