Answer» 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. Hero, the product of three consecutive positive integer is divisible by 6.
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