Question

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

Answer 1

We need to show that, based on that assumption, (k+1)(k+2)(k+3) is also divisible by 6. (k+1)(k+2)(k+3) = (k+1)(k+2)k + (k+1)(k+2)3 = k(k+1)(k+2) + 3(k+1)(k+2). By induction hypothesis, the first term is divisible by 6, and the second term 3(k+1)(k+2) is divisible by 6 because it contains a factor 3 and one of the two consecutive integers k+1 or k+2 is even and thus is divisible by 2. Thus it is divisible by both 3 and 2, which means it is divisible by 6. The theorem is proved since the sum of two multiples of 6 is also a multiple of 6 is also a multiple of 6.