prove that one of every three consecutive posetive integer is divisible by 3
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Answer. Let three consecutive positive integers be n, n + 1 and n + 2. If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3. So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 3.
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