= product of three consecutive positive integers Now, we have to show that the product of three consecutive positive integers is divisible by 6.
We know that any positive integer a is of the form 3q, 3q + 1 or 3q + 2 for some integer q.
Let a, a + 1, a + 2 be any three consecutive integers.
Case I : If a =3q
a(a + 1)(a + 2) =3q(3q + 1)(3q + 2)
=3q (2r)
= 6qr, which is divisible by 6.
(Product of two consecutive integers (3q + 1) and (3q + 2) is an even integer, say 2r)
Case II : If a =3q + 1
∴ a(a + 1)(a + 2) = (3q + 1)(3q + 2)(3q + 3)
=(2r) (3)(q + 1)
=6r(q + 1)
which is divisible by 6.
Case III : If a = 3q + 2
∴ a(a + 1)(a + 2) = (3q + 2)(3q + 3)(3q + 4)
= multiple of 6 for every
q = 6r (say)
which is divisible by 6.
Hence, the product of three consecutive integers is divisible by 6.
