Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How do you prove that


is divisible by 6 by using the method of mathematical induction?

According to my book $$\begin{aligned} (n+1)(n+2)(n+3) &= n(n+1)(n+2)+3(n+1)(n+2)\\ &= 6k + 3*2k'\\ &= 6(k+k')\\ &=6k'' \end{aligned}$$

But I wonder, where does that k come from anyway?

share|cite|improve this question
up vote 3 down vote accepted

The $k$ comes from the induction hypothesis: you are supposed to prove that $(n+1)(n+2)(n+3)$ is divisible by 6 assuming that $n(n+1)(n+2)$ is divisible by 6.

That's from where you get an integer $k$ for which $$n(n+1)(n+2) = 6k.$$

The $k'$ comes from the observation that $(n+1)(n+2)$ is the product of two subsequent numbers. One of which has to be even, so the product is even.

share|cite|improve this answer
Ah, now the k makes sense. The k' still confuses me though. Why to care if a number is even or not? If any number multiplied by 6 instantly becomes divisible by 6 anyway, I think. – Zol Tun Kul Jul 5 '12 at 10:41
The second term in the sum is $3(n+1)(n+2)$. To show that it's divisible by 6 they need to show that $(n+1)(n+2)$ is even and then they substitute $(n+1)(n+2) = 2k'$. – Joni Jul 5 '12 at 10:46
+1 for actually answering the question as asked. Yes, there are easier ways to prove the result without using induction (just observe than one of $n+1$, $n+2$ and $n+3$ must be divisible by 3 and at least one must be even), but that's not what the OP asked about. – Ilmari Karonen Jul 5 '12 at 11:01
It is a bit curious though to use without explanation that $(n+1)(n+2)$ is divisible by $2$ (implicitly checking the possible classes of $n$ modulo $2$), and to not use the very similar argument (checking the possible classes of $n$ modulo $6$) to conclude that $n(n+1)(n+2)$ is divisible by $6$ – Marc van Leeuwen Jul 5 '12 at 11:09
@MarcvanLeeuwen: mostly a tribute to how textbooks aren't very good at coming up with good examples of induction, I suspect :) – Ben Millwood Jul 5 '12 at 11:45

Since one number out of every $2$ consecutive integers is divisible by $2$ and one out of every $3$ consecutive integers is divisible by $3$, therefore, in the product of $3$ consecutive integers $n(n+1)(n+2)$, atleast one number is divisible by $2$ and one divisible by $3$ and $\gcd(2,3)=1 $ $ \implies n(n+1)(n+2)$ is divisible by $2\times3=6$. $$$$ You can also do it this way: Since every integer is one of the forms: $6k,6k+1,6k+2,6k+3,6k+4 $ or $6k+5$, therefore, possibilities the product of three consecutive numbers is : $$6k(6k+1)(6k+2)=6(k(6k+1)(6k+2))$$ $$(6k+1)(6k+2)(6k+3)=(6k+1)2(3k+1)3(2k+1)=6((6k+1)(3k+1)(2k+1))$$ $$(6k+2)(6k+3)(6k+4)=2(3k+1)3(2k+1)(6k+4)=6((3k+1)(2k+1)(6k+4))$$ $$(6k+3)(6k+4)(6k+5)=3(2k+1)2(3k+2)(6k+5)=6((2k+1)(3k+2)(6k+5)).$$ Thus product of every possible combination of $3$ consecutive integers is coming out to be divisible by $6$.

share|cite|improve this answer
I tend to think it unwise to denote multiplication as you did in your first paragraph $2.3=6$. To those of us for whom $.$ is a decimal separator, it looks like a very strange statement indeed... perhaps \times is better? – Ben Millwood Jul 5 '12 at 11:47
@Ben Millwood: Thanks for the suggestion. better now?? – Aang Jul 5 '12 at 11:49
Yep, I'm happy :) – Ben Millwood Jul 5 '12 at 11:49
:):):) me too.. – Aang Jul 5 '12 at 11:50


Let f(n) is divisible by 6 be true for n=m.

f(m+1) = (m+1)(m+2)(m+3)=m(m+1)(m+2) + 3(m+1)(m+2) =f(m) + 3(m+1)(m+2). Now, product of two consecutive number is always even. So, the proposition is true for n = m+1 , if it is true for n=m.

By k means there exists an integer k such that n(n+1)(n+2)/6.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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