For any $a$ in $\Bbb Z$, prove that $6|a(a+5)(a+10)$

Question

So I am given this question for my number theory and proof class:

For any $a \in \Bbb Z$, prove that $6|a(a+5)(a+10)$.

I've thought about a few different ways to approach this. I think I could use the division algorithm, separate into cases, and let $a=6k, 6k+1, 6k+2,...6k+5$ and show that $6|a(a+5)(a+10)$ that way but there has to be a shorter route.

So then I thought I could use the following logic: If $a$ is even, then $a$ is divisible by $2$ (and then I could 'factor' out a $2$). Also, if $a$ is odd, then $(a+5)$ is even and divisible by $2$.

However, I am having trouble proving that one of the terms is a multiple of $3$ using that logic. In the past, I have seen similar proofs to show $6|a(a+1)(a+2)$, but those are consecutive integers and finding the multiples of $2$ and $3$ are much more straightforward.

Any help is appreciated, thank you!

Answer 1

If $a$ is divisible by $3$, done.

If $a + 1$ is divisible by $3$, then so is $a + 10 = a + 1 + 9$.

If $a + 2$ is divisible by $3$, then so is $a + 5 = a + 2 + 3$.

The result follows.

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Alternatively, if you know a little bit of modular arithmetic: $$a(a + 5)(a + 10) \equiv a (a + 2)(a + 1) \pmod{3}$$

Answer 2

Think of it this way: $6 | (a + 0)(a + 5)(a + 10)$; the number on the right has three divisors that we know of. Obviously, $0 \equiv 0 \pmod 3$, and then $10 \equiv 1 \pmod 3$ and $5 \equiv 2 \pmod 3$. So in a way, 0, 10, 5 are "consecutive". There are essentially only three possibilities for $a \equiv n \pmod 3$, but whatever $n$ is, either 0, 10 or 5 will give a multiple of 3.

Along the same lines, either $a + 0$ and $a + 10$ are even, or $a + 5$ is.