# The product of three consecutive natural numbers is divisible by $6$

Please give me feedback for my answer to this question.

Prove or ﬁnd a counterexample: The product of any three consecutive natural numbers is divisible by $6$

My answer: True. Suppose $n$ is a natural number, such that the $3$ consecutive natural numbers is $n, n+1, n+2$. Then, \begin{align}\frac{n(n+1)(n+2)}{6} &= \frac{1(1+1)(1+2)}{6} \\ &= \frac 66 \\ &= 1\end{align} Thus, $n(n+1)(n+2)$ is divisble by 6.

• It looks like all you proved is that it holds for $n=1$... – ec92 Apr 17 '14 at 5:54
• How about if I hold it for all natural numbers? – user142943 Apr 17 '14 at 6:04

## 1 Answer

You have proved that $1\times2\times3$ is divisible by six, not that the product of any 3 consecutive natural numbers is divisible by $6$.

If a number is divisible by $6$, then it must be divisible by both $2$ and $3$. Your product is $$n(n+1)(n+2)$$ so you could try showing that at least one of $n$, $n+1$ or $n+2$ is a multiple of 3, and at least one is even.

• How about this answer? Please give me feedback. – user142943 Apr 17 '14 at 6:55
• My answer: True since if we let 1,2,3 be the 3 consecutive natural number then 1*2*3=6 which 6 is divisble by 6, thus the product of any three consecutive natural number is divisble by 6. – user142943 Apr 17 '14 at 6:56
• You have to prove it for all $n$. You can't just show that 1*2*3 is divisible by 6 because how does that tell you that 2*3*4 is divisible by 6, Or 25*26*27? – David Apr 17 '14 at 7:28
• Following your logic, you could prove that every square number is equal to 1. n^2=1^2=1, so any number to the power of 2 is 1! – David Apr 17 '14 at 7:29