Show that one and only one out of n, (n + 2) and (n + 4) is divisible by 3,

Let q be quotient and r be the remainder.
On applying Euclid's algorithm, i.e. dividing n by 3, we have
n = 3q + r       0 ≤ r < 3
⇒  n = 3q + r       r = 0, 1 or 2
⇒  n = 3q  or  n = (3q + 1) or n = (3q + 2)
Case 1​: If n = 3q, then n is divisible by 3.
Case 2: If n = (3q + 1), then (n + 2) = 3q + 3 = 3(3q + 1), which is clearly divisible by 3.
             In this case, (n + 2) is divisible by 3.
Case 3 : If n = (3q + 2), then (n + 4) = 3q + 6 = 3(q + 2), which is clearly divisible by 3.
              In this case, (n + 4) is divisible by 3.
Hence, one and only one out of n, (n + 1) and (n + 2) is divisible by 3.