Factor the given expression. We have: n³-9n²+20n=n(n-4)(n-5) ··· Ex.1
If Ex.1 is divisible by 6, Ex.1 is also divisible by two prime factors of 6, 2 and 3. (6=2x3)
A. If n is odd, (n-5) is even. If n is even, (n-4) is also even. Therefore, n(n-4)(n-5) is always a multiple of 2.
B. If Exp.1 is divisible by 3, the remainder is 0,1,or 2.
If n ≡ 0 (mod 3), n-4 ≡ -4 ≡ -1 (mod 3), and n-5 ≡ -5 ≡ -2 (mod 3), that is: n is a multiple of 3.
If n ≡ 1 (mod 3), n-4 ≡ -3 ≡ 0 (mod 3), and n-5 ≡ -4 ≡ -1 (mod 3), that is: (n-4) is a multiple of 3.
If n ≡ 2 (mod 3), n-4 ≡ -2 (mod 3), and n-5 ≡ -3 ≡ 0 (mod 3), that is: (n-5) is a multiple of 3.
Therefore, n(n-4)(n-5) is always a multiple of 3.
CK: If n=1, n(n-4)(n-5)=1(-3)(-4)=12, 12(n=2), 6(n=3), 0(n=4), 0(n=5), 12, 42, 96, 180 … CKD.
Therefore, n³-9n²+20n is divisible by 6 for all integers greater than or equal to 1.