Is factorial divisibility rule with cons. multiples?

[whua88]

Rule: The product of k consecutive integers is always divisible by k factorial (k!).
(Guide 1, Chapter 4 - Consecutive Integers, Page 31)

In this section, in order to prove this rule, the chapter asks us at first to come up with a series of 3 consecutive integers in which none of the integers is a multiple of 3 and then lists a string of examples like:

1 x 2 x 3 = 6
2 x 3 x 4 = 24
3 x 4 x 5 = 60
4 x 5 x 6 = 120

Does this rule also hold true for consecutive multiples?

[jnelson0612]

That's a great question! Let's take a look at a few:

Multiples of 3: 3 X 6 X 9 = a number that will be divisible by 3! (since 3! is 6)
6 X 9 X 12=divisible by 3!
and so on

Let's try 4:
4 X 8 X 12 X 16 = a number that will be divisible by 4 * 3 * 2 * 1 (we can see that we have a 4, we have 12 which has a factor of 3, and we have two other even numbers so that takes care of the 2)

Let's try 5:
5 X 10 X 15 X 20 X 25
Will be divisible by 5, 4 (factor of 20), 3 (factor of 15), 2 (factor of 10 and 20)

Thus, that looks correct!