Let the three consecutive numbers be 3n-1, 3n,3n+1

Then the product of the three numbers will be

(3n-1)\cdot3n\cdot (3n+1)

=3n(9n^2-1)

Obviously 3n(9n^2-1) is divisible by 3 and it's also an even number. Why?

if n is an even number, our claim is true.

If n is an odd number, then n^2 is also an odd number. 9 is odd. The product of two odd numbers is still an odd number. However, if subtract 1 from the odd number, then the difference is even.

So in both cases, 3n(9n^2-1) is even.

Therefore, the product of three consecutive integers is divisible by 6.