2

Prove that the product of three consecutive positive integers is divisible by 6 by expressing the positive integer n as n=8*q+r

I expressed the problem as n(n+1)(n+2) where n is a positive integer I expressed n as n=6*q+r using Euclid's lemma, where r={0,1,2,3,4,5} and I proved n(n+1)(n+2) is a multiple of 6. Now the question is why should we start with the premises that n=6*q+r, why not by expressing it as n=8*q+r as n(n+1)n+2)=8*q(8*q+1)(8*q+2)

Achari S Ganesha
  • 908
  • A good question! I really don't see why one would start working on this problem by writing $n=8q+r$. That won't allow you to identify the factor that is divisible by three, so it is not a helpful hint. $n=6q+r$ is, indeed, "what the doctor orders" to cure this problem. – Jyrki Lahtonen Jul 08 '14 at 07:47

3 Answers3

2

Hint: $6|a \iff 2|a \ \& \ 3|b$.

Mohammad Khosravi
  • 2,363
1

Another way: Out of any three consecutive numbers, one is necessarily even, and one is divisible by 3 ; by, e.g., piegeonhole principle on congruence clases: 3 different classes 3, numbers.

user99680
  • 6,836
  • 16
  • 25
1

Let's see the different cases in $\mathbb Z_6$:

  • $\overline 0\times\overline 1\times\overline 2$
  • $\overline 1\times\overline 2\times\overline 3$
  • $\overline 2\times\overline 3\times\overline 4$
  • $\overline 3\times\overline 4\times\overline 5$
  • $\overline 4\times\overline 5\times\overline 6$
  • $\overline 5\times\overline 6\times\overline 1$ and always the result is $\overline 0$. Conclude.