Is positive integer n – 1 a multiple of 3?

[shaw.s.li]

Is positive integer n - 1 a multiple of 3?

(1) n^3 - n is a multiple of 3

(2) n^3 + 2n^2+ n is a multiple of 3

DS question:

1. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

2. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

3. Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient.

4. EACH statement ALONE is sufficient.

5. Statements (1) and (2) TOGETHER are NOT sufficient.

My question is in regards to simplifying statement (1): n^3 - n is a multiple of 3. I understand that n^3-n can be simplified to (n-1)(n)(n+1). It is stated in the book that for any three consecutive integers, there is always one multiple of 3. However, is this rule true for only 3 consecutive integers?

For example, if I have 4 consecutive integers, isn't the rule that there will always be 1 multiple of 4 because (n-1)(n)(n+1)(n+2)

If n=even, then n and n+2 are both even, thus two even numbers means two multiples of 2 or 1 multiple of 4.

If n=odd, then n-1 and n+1 are both even. Same logic as above.

Then, in addition to the rule that the product of k consecutive integers is always divisible by k!, shouldn't the rule also state that the product of k consecutive integer always contains one multiple of k?

I'm trying to understand this rule so that I can apply it correctly.

[tim]

you are correct. good job..