Prove that one of every three consecutive positive integers is divisible by $3$.

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Step 1: Let $n$ be any positive integer. Divide it by $3$.
$n = 3q + r$, $r = 0, 1, 2$
Step 2: Case 1: $n = 3q$ (divisible by 3)
Case 2: $n = 3q + 1 \Rightarrow n + 2 = 3q + 3 = 3(q+1)$ (divisible by 3)
Case 3: $n = 3q + 2 \Rightarrow n + 1 = 3q + 3 = 3(q+1)$ (divisible by 3)
Step 3: Hence one of every three consecutive positive integers is divisible by $3$.

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