Prove that $n(n+1)(n+5)$ is a multiple of $6$ I need to prove that $n(n+1)(n+5)$ is divisible by 6. where $n$ is a natural number. I have used the method of induction. But not successful
I got the expression $(k^3+6k^2+5k)+3k^2+15k+12$ when $n=k+1$. 
The term inside the bracket is divisible by 6 since we have assumed that the result is true when $n=k$. If we can show that $3k^2+15k+12$ is also divisible by 6, then we are done. But how to proceed? 
 A: Since we can easily see that $3k^2+15k+12=3(k^2+5k+4)$ is divisible by $3$, it remains to show that it is divisible by $2$.
It suffices to look at $$k^2+k=3k^2+15k+12-2(k^2+7k+6)$$and this is even because $k^2+k=k(k+1)$ is the product of two consecutive natural numbers, and therefore divisible by $2$.
You may also go deeper and prove that $k^2+k$ is even using induction, of course. It will then be induction-within-an-induction.
A: $$
\begin{align}
n(n+1)(n+5)
&=n(n+1)(n+2)+3n(n+1)\\[6pt]
&=6\binom{n+2}{3}+6\binom{n+1}{2}
\end{align}
$$

Binomial Coefficient Basics
If, instead of Pascal's Triangle, we define the binomial coefficients as
$$
\binom{n}{k}=\frac{n(n-1)\cdots(n-k+1)}{k!}
$$
Then we have
$$
\begin{align}
\binom{n-1}{k}+\binom{n-1}{k-1}
&=\binom{n-1}{k-1}\frac{n-k}k+\binom{n-1}{k-1}\\
&=\binom{n-1}{k-1}\frac nk\\
&=\binom{n}{k}
\end{align}
$$
Thus, if $\binom{n-1}{k}\in\mathbb{Z}$ for all $k\in\mathbb{Z}$, then $\binom{n}{k}\in\mathbb{Z}$ for all $k\in\mathbb{Z}$.
A: $$n(n+1)(n+5)=n^3+6n^2+5n\equiv n^3-n\pmod6$$
Now $n^3-n=(n-1)n(n+1)$ 
and use The product of n consecutive integers is divisible by n factorial
A: One number among $n$, $n+1$, $n+2$ is a multiple of 3.
Since $5 \pmod 3 \equiv 2$, among $n$, $n+1$ and $n+5$, one is a multiple of 3.
Either $n$ or $n+1$ must be an even number (since they are consecutive numbers).
Therefore, n(n+1)(n+5) is a multiple of 2.
As $n(n+1)(n+5)$ is a has both 3 and 2 as factors, it is a multiple of 6. 
A: Continuing from your start, let $k^3+6k^2+5k=6p$ for some $p\in\mathbb Z$.
Now, 
$$(k+1)(k+2)(k+6)=(k^3+6k^2+5k)+3k^2+15k+12=6p+3k^2+15k+12=6p+3(k^2+5k+4)=6p+3(k+1)(k+4)$$

Case 1: $k$ is even.
Then, $k+4$ is also even. We can thus write $k+4=2q$ for some $q\in\mathbb Z$.
Hence,
$$6p+3(k+1)(k+4)=6p+3(k+1)(2q)=6p+6q(k+1)=6(p+q(k+1))$$
where $(p+q(k+1))\in\mathbb Z$.
Thus, the statement is true when $k$ is even.

Case 2: $k$ is odd.
Then, $k+1$ is  even. We can thus write $k+1=2q$ for some $q\in\mathbb Z$.
Hence,
$$6p+3(k+1)(k+4)=6p+3(k+4)(2q)=6p+6q(k+4)=6(p+q(k+4))$$
where $(p+q(k+4))\in\mathbb Z$.
Thus, the statement is true when $k$ is odd.
A: Suppose $n$ is a multiple of 6. Then we are done. Suppose $n$ is neither a multiple of 2 or 3. Then $n+1$ must be even, and if $n+1$ is still not a multiple of 3 then $n+5$ is. Note $n+5\equiv n+2$ (mod 3) and since neither $n$ nor $n+1$ is a multiple of 3, then $n+2$ is. If $n$ is even but not a multiple of 3, we proceed the same way to show either $n+1$ or $n+2$ is. The only case left is when $n$ is an odd multiple of 3 then $n+1$ is even and we are done. In any case, we get a factor of 2 and a factor of 3.
A: You are almost done! Take $3k^2+15k+12$ modulo $6$ to reduce it to proving that $3k(k+1)$ is divisible by $6$. We already have a factor of $3$, so it suffices to show that $k(k+1)$ must be even. But out of any two consecutive numbers exactly one must be even, so their product must be even too; we are done.
A: You just needed one more step.
$$\begin{align}n(n+1)(n+5) = & (n+1)(n+2)(n+6) \\ = & n^3+9 n^2+20 n+12 \\ = & (n^3+6n^2+5n)+3n^2+15n+12 \\ = & n(n+1)(n+5)+\underbrace{3n(n+5)}_{\star}+12\end{align}$$
$\bigstar$ If $n$ is even, then $3n$ is divisible by 6, otherwise $n$ is odd and $3(n+5)$ is divisible by 6.

Of course you could have done this immediately.


*

*$n(n+1)$ is divisible by $2$.

*If neither $n$ nor $n+1$ are divisible by $3$, then $n+2$ and $n+5$ will be.


*

*exactly one of the three factors, $n$, $n+1$, $n+5$ is divisible by $3$.


*Therefore $n(n+1)(n+5)$ is divisible by $2$ and $3$, and thus by $6$.

