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Question
Prove that the product of three consecutive positive integer is divisible by 6.
Solution
Let, n be any positive integer. Since any positive integer is of the form 6q or 6q + 1 or 6q + 2 or, 6q + 3 or 6q + 4 or 6q + 5.
If n = 6q, then
n(n + 1)(n + 2) = (6q + 1)(6q + 2)(6q + 3)
= 6[(6q + 1)(3q + 1)(2q + 1)]
= 6m, which is divisible by 6?
If n = 6q + 1, then
n(n + 1)(n + 2) = (6q + 2)(6q + 3)(6q + 4)
= 6[(6q + 1)(3q + 1)(2q + 1)]
= 6m, which is divisible by 6
If n = 6q + 2, then
n(n + 1)(n + 2) = (6q + 2)(6q + 3)(6q + 4)
= 6[(3q + 1)(2q + 1)(6q + 4)]
= 6m, which is divisible by 6.
If n = 6q + 3, then
n(n + 1)(n + 2) = (6q + 3)(6q + 4)(6q + 5)
= 6[(6q + 1)(3q + 2)(2q + 5)]
= 6m, which is divisible by 6.
If n = 6q + 4, then
n(n + 1)(n + 2) = (6q + 4)(6q + 5)(6q + 6)
= 6[(6q + 4)(3q + 5)(2q + 1)]
= 6m, which is divisible by 6.
If n = 6q + 5, then
n(n + 1)(n + 2) = (6q + 5)(6q + 6)(6q + 7)
= 6[(6q + 5)(q + 1)(6q + 7)]
= 6m, which is divisible by 6.
Hence, the product of three consecutive positive integer is divisible by 6.
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