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The objective is to prove that $n(n+1)(n+5)$ is a multiple of 3.

I took the following simplistic route $$n(n+1)(n+5) = 3X$$ $$n(n+1)(n+5)\frac{n+2}{n}\frac{n+6}{n+5} = 3X*\frac{n+2}{n}\frac{n+6}{n+5}$$

Now, since $(n+1)(n+2)(n+6) = 3X*Y$, n+1 should be divisible by 3 as long as n is divisible by 3 (Proof by induction).

Obviously i don't feel that is the right answer, it feels like cheating. I can't see how this method is illegal however.

Marc van Leeuwen
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Aayush Agrawal
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  • The quantity you are referring as $Y$, why is that an integer? – Anurag A Jan 30 '16 at 06:30
  • Since this is a proof by induction, i assume that Y is an integer since the assumption is that n is a multiple of 3. – Aayush Agrawal Jan 30 '16 at 06:30
  • Are you given that $n$ is a multiple of $3$? – Anurag A Jan 30 '16 at 06:34
  • No. The objective is to prove $n(n+1)(n+5)$ is a multiple of 3 by induction. In it, we're assuming that any natural number n is a multiple of 3 and then applying the induction case to prove that if n is a multiple of 3, so must n+1. So, it is not given that n is a multiple of 3, but it is an assumption for the induction. – Aayush Agrawal Jan 30 '16 at 06:35
  • That is an incorrect assumption. How can $n$ and $n+1$ both be multiples of $3$. Unless you meant that you are assuming the statement holds for $n$ case and you are trying to prove for the $n+1$ case. – Anurag A Jan 30 '16 at 06:38
  • I made a mistake in the comment above. We're trying to prove that $n(n+1)(n+5)$ is a multiple of 3 for all natural numbers n. – Aayush Agrawal Jan 30 '16 at 06:39
  • Related : http://math.stackexchange.com/questions/971367/use-the-division-algorithm-to-prove-that-3n%C2%B3-2n-for-all-n-%E2%88%88-%E2%84%95 – lab bhattacharjee Jan 30 '16 at 06:43
  • Well, of course n(n+1)(n+5) is divisible by 3 if n is divisible by three. That's obvious. But you have to prove n(n+1)(n+5) is divisible by 3 if n is not divisible by three. – fleablood Jan 30 '16 at 07:32
  • "since the assumption is that n is a multiple of 3" "we're assuming that any natural number n is a multiple of 3". why on earth do you think that is what you are supposed to be doing? You know that that is not true because 1,2, 4,5,7,8,10 etc. are all not multiples of 3. So assuming that any natural number n is a multiple of 3, is a ridiculous and pointless thing to assume I don't think you understand how a proof by induction is supposed to work. – fleablood Jan 30 '16 at 07:36
  • Prove by induction: show that if n = 1; n(n+1)(n+5) = 126 = 12 is divisible by 3. show that if we somehow knew that if for some m, m(m+1)(m+5) is divisible by 3, then we can prove that (m+1)(m+2)(m+6) is divisible by 3. Note: we can do that: m(m+1)(m+5) is div by 3 so (m+3k)(m+1 + 3j)(m+ 5 + 3v) are also divisible by 3 for all k,j,v so (m+6)(m+1)(m + 5 -3) = (m+1)(m+2)(m+6). – fleablood Jan 30 '16 at 07:43

5 Answers5

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Hint: $n(n+1)(n+5) = n(n+1)(n+2 + 3)$

$= n(n+1)(n+2)+3n(n+1)$

So just prove the product of 3 consecutive numbers is a multiple of 3.

Now as Anurag mentioned, $n$ can be of the form $3k$ or $3k+1$ or $3k+2$. In all cases, show that $n(n+1)(n+2)$ is divisible by 3.

Max Payne
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  • I'm sorry but i don't understand. Why is (n+1)(n+2+3) consecutive? – Aayush Agrawal Jan 30 '16 at 06:31
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    @AayushAgrawal what Tim means is $n+2$ and $n+2+3$ will have the same remainder when divided by $3$. – Anurag A Jan 30 '16 at 06:35
  • @AayushAgrawal Edited. I am not saying $(n)(n+1)(n+2+3)$ are consecutive. see edit – Max Payne Jan 30 '16 at 06:40
  • But this isn't a proof by induction. Although it is better in my opinion. Although the proof by induction has a very nice twist in it: 3 | n(n+1)(n+5) iff 3 | (n+6)(n+1)(n + 2) = (n+1)(n+2)(n+6). – fleablood Jan 30 '16 at 07:46
  • @fleablood True! I didn't see the post properly! Induction proof( as you marty and anurag too posted) is concise, and doesn't need cases. – Max Payne Jan 30 '16 at 08:07
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hint

For any integer $n$ there are three possibilities: either $n=3k$ or $n=3k+1$ or $n=3k+2$. For each such possibility show that one of the terms in the product is a multiple of $3$.

For proceeding with induction: assume it holds for the $n$ case, i.e assume that $$n(n+1)(n+5)=\color{blue}{n^3+6n^2+5n}=3k$$

Now consider \begin{align*} (n+1)(n+2)(n+6)&=n^3+9n^2+20n+12\\ &=\color{blue}{(n^3+6n^2+5n)}+3n^2+15n+12\\ &=3k+3(n^2+5n+4). \end{align*}

Thus the left hand side is a multiple of $3$.

Anurag A
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Your proof is not correct because $\frac{n+2}{n}\frac{n+6}{n+5} $ is never an integer. Just looking at it, $\frac{n+2}{n}\frac{n+6}{n+5} =\frac{n^2+8n+12}{n^2+5n} =\frac{n^2+5n+3n+12}{n^2+5n} =1+\frac{3n+12}{n^2+5n} $, and $n^2+5n > 3n+12$ when $n^2+2n>12$ or $(n+1)^2 > 13$ which is true for $n \ge 3$.

A proof by induction could go like this:

Base case: For $n=1$, $n(n+1)(n+5) =1\cdot 2 \cdot 6 =12 $ is divisible by 3.

Induction step: The difference between the expression for $n+1$ and $n$ is

$\begin{array}\\ (n+1)(n+2)(n+6)-n(n+1)(n+5) &=(n^3+9 n^2+20 n+12)-(n^3+6n^2+5n)\\ &=3n^2+15n+12\\ &=3(n^2+5n+4) \end{array} $

and this is obviously divisible by 3.

Therefore, if $n(n+1)(n+5)$ is divisible by 3, so is $(n+1)(n+2)(n+6)$.

Since $n(n+1)(n+5)$ is divisible by 3 for $n=1$, it is divisible by 3 for all $n$.

marty cohen
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The proof by induction is !!CUTE!!!

Base: step. If $n = 1$ then $n(n+1)(n+5) = 1*2*6 = 12$ is divisible by 3.

Induction: Assume that for some $m$ we know that $m(m+1)(m+5)$ is divisible by 3, we want to prove that $(m+1)(m + 1 + 1)(m + 1 + 5)$ is divisible by 3.

Proof:

Let $m(m+1)(m+5) = 3X$.

$(m+1)(m+1+1)(m + 1 + 5) = $(m+1)(m+2)(m+6)$

$=(m+1)(m + 2 + 3)(m+6) - 3[(m+1)(m+6)]$

$= (m+1)(m + 5)m + 6[(m+1)(m+5)] - 3[(m+1)(m+6)]$

$= 3X + 6[(m+1)(m+5)] - 3[(m+1)(m+6)]$

$= 3[X + 2[(m+1)(m+5)] -[(m+1)(m+6)]$ is divisible by 3.

So we are done.

By induction we not the proposition is true for n = 1. we know if it is true for n it is true for n+1, so inductively it must be true for all natural numbers.

fleablood
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If you're looking for a proof by induction, start by labeling $f(n)=n(n+1)(n+5)$. Then, you can show that $f(n+1)-f(n)=3(n+1)(n+4)$, which is a multiple of $3$. So:

  1. $f(0)=0$ is a multiple of $3$
  2. If $f(n)$ is a multiple of $3$, then $f(n+1)=[f(n)]+[f(n+1)-f(n)]$ is a sum of two multiples of $3$, and is hence a multiple of $3$ itself.

Induction does the rest of the work for us from here.

Perhaps interesting to note that $f(n)$ is always a multiple of $6$ - this takes slightly more work, but can be adapted readily by showing that $3(n+1)(n+4)$ is always a multiple of $6$.

πr8
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