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Question
Prove that one of every three consecutive positive integers is divisible by 3.
Theorem
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Solution
Let n be any positive integer.
∴ n = 3q + r, where r = 0, 1, 2
Putting r = 0,
n = 3q + 0 = 3q, which is divisible by 3.
Putting r = 1,
n = 3q + 1, which is not divisible by 3.
Putting r = 2,
n = 3q + 2, which is not divisible by 3.
Hence, one of every three consecutive positive integers is divisible by 3.
Hence Proved.
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