What is the greatest whole number that must be a divisor of the product of any three consecutive positive integers?

6

Step-by-step explanation:

We can establish 6 as an upper bound, since 1*2*3 = 6, and 6 is clearly the greatest number that is a divisor of itself.

We can show that the product of any three consecutive numbers is divisible by 6, because out of any three consecutive integers, at least one must be divisible by 3, and at least one must be divisible by 2. Since the product must have factors of 2 and 3, it must also have 6 as a factor.

3000000000000000000000

i'll be glad to : )