Prove that one of every three consecutive positive integer is divisible by 3

Hello:

all n in N; n (n+1) (n+2) = 3a a in N or : ≡ 0 (mod 3)

1) n ≡ 0 (mod 3) ... (1)

n+1 ≡ 1 (mod 3) ... (2)

n+2 ≡ 2 (mod 3) ... (3)

by (1), (2), (3) : n (n+1) (n+2) ≡ 0*1*2 (mod 3) : ≡ 0 (mod 3)

2) n ≡ 1 (mod 3) ... (1)

n+1 ≡ 2 (mod 3) ... (2)

n+2 ≡ 3 (mod 3) ... (3)

by (1), (2), (3) : n (n+1) (n+2) ≡ 1*2 * 3 (mod 3) : ≡ 0 (mod 3), 6≡ 0 (mod)

3) n ≡ 2 (mod 3) ... (1)

n+1 ≡ 3 (mod 3) ... (2)

n+2 ≡ 4 (mod 3) ... (3)

by (1), (2), (3) : n (n+1) (n+2) ≡ 2*3 * 4 (mod 3) : ≡ 0 (mod 3), 24≡ 0 (mod3)