6
$\begingroup$

Prove that $n(n+1)(n+2)$ is divisible by $6$ for all integers.

I'm not sure if I'm suppose to use division into cases or not. Our teacher ran out of time to go over this in class, and this is on my homework.

$\endgroup$
3
  • $\begingroup$ Unless you’re specifically required to prove it by induction (which in my opinion would be poor pædagogy), I wouldn’t: I’d show that $n(n+1)(n+2)$ has at least one factor of $2$ and at least one factor of $3$. $\endgroup$ – Brian M. Scott Sep 23 '13 at 0:37
  • $\begingroup$ You got consecutive integers unless one of the factors is 0 you will have a even number so you should be able to pull out a 2 and another factor will be a multiple of 3 so you can pull out a 3. $\endgroup$ – user60887 Sep 23 '13 at 1:05
  • $\begingroup$ If you understand 2, 3 should easily follow. Pick a number, any number. Now, add 1, then add two. You see how one of these three must be a factor of three? (4,5,6) has the 6, (6,7,8) still 6. (7,8,9) there's a 9. $\endgroup$ – JTP - Apologise to Monica Sep 23 '13 at 4:19
6
$\begingroup$

An integer is divisible by $6$ if and only if it is divisible by both $2$ and $3$. Can you see why $n(n+1)(n+2)$, the product of three consecutive integers, is divisible by $2$? What about by $3$?

$\endgroup$
3
  • $\begingroup$ I see that because they are consecutive integers, one of the integers must be divisible by two, and one must be divisible by 3, but as far as being able to translate that into a proof, I'm having trouble with that $\endgroup$ – Larry Sep 23 '13 at 0:56
  • $\begingroup$ What you said is pretty much sufficient. Given $k$ consecutive integers, precisely one of them is divisible by $k$. If you like, you can prove this result which will give you what you need. $\endgroup$ – Michael Albanese Sep 23 '13 at 1:02
  • $\begingroup$ Proving such a result almost surely requires some use of the Euclidean algorithm, which is proved with induction. When learning something as important as induction, it is nice to see some concrete examples that can be proved directly from induction (this question is such an example). $\endgroup$ – PVAL-inactive Sep 23 '13 at 8:39
4
$\begingroup$

I'd recommend trying this simpler question first:

Prove that $n(n+1)$ is divisible by $2$ (i.e., is even) for all integers $n$.

Note that if either $n$ or $n+1$ is even, then $n(n+1)$ will be even. Must either $n$ or $n+1$ be even?

Once you've solved this problem, the generalization to your original problem should be clear.

$\endgroup$
2
$\begingroup$

Big Hint: $n(n+1)(n+2)-(n-1)(n)(n+1)=3(n^2+n)$

$\endgroup$
4
  • $\begingroup$ I don't see how this is a big hint. $\endgroup$ – Michael Albanese Sep 23 '13 at 1:15
  • $\begingroup$ factorise again $3n(n+1)$, if n is odd n+1 is... and vice versa $\endgroup$ – sachinruk Sep 23 '13 at 1:29
  • $\begingroup$ @Sachin_ruk: How does that help you solve the problem? $\endgroup$ – Michael Albanese Sep 23 '13 at 8:18
  • $\begingroup$ @MichaelAlbanese Apply the title of the question... $\endgroup$ – PVAL-inactive Sep 23 '13 at 8:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.