Prove that if a positive number is of the form 6q+5, then it is of the form 3q+2 for some integer, but not conversely

Let n be any positive integer.

Hence, the three consecutive positive integers are n, n+1, and n+2.

We know that any positive integer can be of the form 6q, or 6q+1, or 6q+2, or 6q+3, or 6q+4, or 6q+5. {From Euclid’s division lemma for b= 6}

ThereforeFor n= 6q,

⇒ n(n+1)(n+2)= 6q(6q+1)(6q+2)

= 6[q(6q+1)(6q+2)]

= 6m, which is divisible by 6. [m= q(6q+1)(6q+2)]

For n= 6q+1,

⇒ n(n+1)(n+2)= (6q+1)(6q+2)(6q+3)

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

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

For n= 6q+2,

⇒ n(n+1)(n+2)= (6q+2)(6q+3)(6q+4)

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

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

For n= 6q+3,

⇒ n(n+1)(n+2)= (6q+3)(6q+4)(6q+5)

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

= 6m, which is divisible by 6. [m= (2q+1)(3q+2)(6q+5)]

For n= 6q+4,

⇒ n(n+1)(n+2)= (6q+4)(6q+5)(6q+6)

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

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

For n= 6q+5,

⇒ n(n+1)(n+2)= (6q+5)(6q+6)(6q+7)

6[(6q+5)(q+1)(6q+7)]

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

Hence, the product of three consecutive positive integers is divisible by 6.

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