nicok06 wrote:

Can I conclude that all the above stated factors are included in the set

Yes, you can - you don't want to eliminate the overlaps. The general rule is this: the product of k consecutive positive integers is always divisible by k!

So, for example, if you multiply six consecutive integers, the product will always be divisible by 6! = 6*5*4*3*2*1, and of course by any factor of 6! as well.

The reason this is all true is because multiples are equally spaced - since multiples of 6 are 6 apart, then in any list of six consecutive integers, we must always have exactly one multiple of 6. Similarly, since multiples of 5 are 5 apart, we always must have at least one multiple of 5 (and possibly two of them) among any six consecutive integers, and so on.

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