product of three consecutive integer is divisible by 6 : Is it true ? Why ?
Answers
Let n, n+1, n+2 be the three consecutive positive integer. We know that any positive integer is of the 6q,6q+1, 6q+2, 6q+3, 6q+4 or 6q+5.
If n=6q, then
n(n+1)(n+2)=6q(6q+1)(6q+2), which is divisible by 6.
If n=6q+1, then
n(n+1)(n+2)=(6q+1)(6q+2)(6q+3)
{Taking common outside}
n(n+1)(n+2)=6(6q+1)(3q+2)(2q+1), which is divisible by 6.
If n=6q+2, then
n(n+1)(n+2)=(6q+2)(6q+3)(6q+4)
=12(3q+1)(2q+1)(3q+2), which is divisible by 6.
If n=6q+3, then
n(n+1)(n+2)=(6q+3)(6q+4)(6q+5)
=6(2q+1)(3q+2)(6q+5), which is divisible by 6.
If n=6q+4, then
n(n+1)(n+2)=(6q+4)(6q+5)(6q+6)
=12(3q+2)(6q+5)(q+1), which is again 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) which is also divisible by 6.