Prove that (n-2)(n-1)(2n-3) is divisible by 6 for any positive integer greater than 2. Where did I go wrong?

P(n): (n - 2)(n - 1)(2n - 3) is divisible by 6 for any positive integer greater than 2.

When n = 3 :
(3 - 2)(3 - 1)(2*3 - 3) = 6
Hence, P(3) is true.

Suppose P(k) is true, i.e. (k - 2)(k - 1)(2k - 3) = 6Q, where Q is a positive integer.
Prove that P(k + 1) is also true.

Proof:
When n = k + 1:
[(k + 1) - 2] [(k + 1) - 1] [2(k + 1) - 3]
= (k - 1)k(2k - 1)
= (k - 1)(2k² - k)
= (k - 1)[(2k² - 7k + 6) - (-7k + 6) - k]
= (k - 1)[(2k² - 7k + 6) + (6k - 6)]
= (k - 1)[(k - 2)(2k - 3) + 6(k - 1)]
= (k - 1)(k - 2)(2k - 3) + 6(k - 1)²
= 6Q + 6(k - 1)²
= 6 * [Q + (k - 1)²]
Hence, P(k + 1) is true.

P(3) is true.
P(k + 1) is true when P(k) is true.
According to the principle of Mathematical Induction, P(n) is true.

("mod 6" can not be used to solve this question.)

n-1 and n-2 are consecutive positive integers, therefore 1 of them is even and the other 1 is odd. The product of three integers is an even number if one of the integers is even, so this proves the product is even. If neither n-1 nor n-2 is divisible by 3, then n is divisible by 3, since n, n-1, and n-2 are 3 consecutive integers. If n is divisible by 3, so is (2n-3). Thus of the 3 factors of the product (n-1)(n-2)(2n-3) at least 1 factor is even and at least 1 factor is divisible by 3. Therefore the product is divisible by 6.