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Prove that the product of three consecutive positive integers is divisible by 6.

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Let a be any positive integer and b = 6.

By division lemma there exists integers q and r such that

a = 6q+r where 0 ≤ r < 6a = 6q , or a = 6q+1 or, a = 6q+2 or, a = 6q+3 or a = 6q+4 or a = 6q+5

Let n is 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.

Case 1:

If n = 6q then

n(n + 1)(n + 2) = 6q (6q + 1)(6q + 2), which is divisible by 6.

Case 2:

If n = 6q + 1, then

n(n + 1)(n + 2) = (6q + 1)(6q + 2) (6q + 3)

= (6q + 1) 2(3q + 1) 3(2q + 1)

= 6(6q + 1)(3q + 1)(2q + 1), which is divisible by 6.

Case 3:

If n = 6q + 2, then

n(n + 1)(n + 2) = (6q + 2)(6q + 3)(6q + 4)

= 2(3q + 1) 3(2q + 1)(6q + 4)

= 6(3q + 1)(2q + 1)(6q + 4), which is divisible by 6.

Case 4:

If n = 6q + 3, then

n(n + 1)(n + 2) = (6q + 3) (6q + 4)(6q + 5)

= 3(2q + 1)2(3q + 2)(6q + 5)

= 6(2q + 1)(3q + 2)(6q + 5), which is divisible by 6

Case 5:

If n = 6q + 4, then

n(n + 1)(n + 2) = (6q + 4) (6q + 5)(6q + 6)

= (6q + 4) (6q + 5) 6(q + 1)

= 6(6q + 4) (6q + 5)(q + 1), which is divisible by 6.

Case 6:

If n = 6q + 5, then

n(n + 1)(n + 2) = (6q + 5) (6q + 6)(6q + 7)

= (6q + 5) 6(q + 1)(6q + 7)

= 6(6q + 5) (q + 1)(6q + 7), which is divisible by 6.

Hence,

n(n + 1)(n + 2) is divisible by 6

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