proof;by method (1)Let n be any positive integer.and the three consecutive positive integers are n, n+1 and n+2. From Euclid’s division lemma for b= 6, 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. So, Case(1) If n= 6q, then, n(n+1)(n+2)= 6q(6q+1)(6q+2) ⇒ n(n+1)(n+2)= 6[q(6q+1)(6q+2)] ⇒ n(n+1)(n+2)= 6m, which is divisible by 6. ......................here,[m= q(6q+1)(6q+2)] Case(2) If, n= 6q+1, then, n(n+1)(n+2)= (6q+1)(6q+2)(6q+3) ⇒ n(n+1)(n+2)= 6[(6q+1)(3q+1)(2q+1)] ⇒ n(n+1)(n+2)= 6m, which is divisible by 6...............here, [m= (6q+1)(3q+1)(2q+1)] Case (3) If n= 6q+2, then, n(n+1)(n+2)= (6q+2)(6q+3)(6q+4) ⇒ n(n+1)(n+2)= 6[(3q+1)(2q+1)(6q+4)] ⇒ n(n+1)(n+2)= 6m which is divisible by 6...............here, [m= (3q+1)(2q+1)(6q+4)] Case(4) If, n= 6q+3, then, n(n+1)(n+2)= (6q+3)(6q+4)(6q+5) ⇒ n(n+1)(n+2)= 6[(2q+1)(3q+2)(6q+5)] ⇒ n(n+1)(n+2)= 6m, which is divisible by 6..................here, [m= (2q+1)(3q+2)(6q+5)] Case(5) If, n= 6q+4, then, n(n+1)(n+2)= (6q+4)(6q+5)(6q+6) ⇒ n(n+1)(n+2)= 6[(6q+4)(6q+5)(q+1)] ⇒ n(n+1)(n+2)= 6m, which is divisible by 6. .......... here, [m= (6q+4)(6q+5)(q+1)] Case(6) If, n= 6q+5, then n(n+1)(n+2)= (6q+5)(6q+6)(6q+7) ⇒ n(n+1)(n+2)= 6[(6q+5)(q+1)(6q+7)] ⇒ n(n+1)(n+2)= 6m, which is divisible by 6. [m= (6q+5)(q+1)(6q+7)] Hence, the product of three consecutive positive integers is always divisible by 6. |
Proof; by method (2) Let n
be any positive integer. and the three consecutive positive integers are n, n+1
and n+2. we know that any positive integer can be of the form 3q, or 3q+1, or 3q+2.. Therefore, n=3q or 3q+1 or 3q+2, where q is some integer. If n=3p, then n is divisible by 3. If n=3p+1, then n+2=3p+1+2=3p+3 =3(p+1) is divisible by 3. here n+2 is divisible by 3 If n=3p+2, then n+1=3p+2+1=3p+3 =3(p+1) is divisible by 3. here n+1 is divisible by 3 So, we can say that one of the numbers among n,n+1 and n+2 is always divisible by 3 , thus the product of these three n, n+1 and n+2 is always divisible by 3 Similarly, From Euclid’s division lemma for b= 2, we know that any positive integer can be of the form 2q, or 2q+1. Therefore, n=2q or 2q+1, where q is some integer. If n=2q, then n and n+2=2q+2=2(q+1) are divisible by 2. If n=2q+1, then n+1=2q+1+1=2q+2=2(q+1) is divisible by 2. So, we can say that one of the numbers among n, n+1 and n+2 is always divisible by 2, thus the product of these three n, n+1 and n+2 is always divisible by 2. Since, n(n+1)(n+2) is divisible by 2 and 3 both. Hence, n(n+1)(n+2) will be always divisible by 6. |
