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)

The answer is b i’m sure