Question

If \(a\) is an integer, then \(a\), \(a + 2\) OR \(a + 4\) must be divisible by 3. Proof?

Answer 1

Is this a case-by-case proof?

Answer 2

Let \(a=3k, k\in \mathbb{Z}\). Then a is divisible by 3.
Let \(a=3k+1, k \in \mathbb{Z}\) Then \(amod3=(3k+1)mod3=1mod3\). Then \((a+2)mod3=(1+2)mod3=3mod3=0mod3\). Hence a+2 is divisible by 3.
Let \(a=3k+2, k \in \mathbb{Z}\). Then \(amod3=2mod3\). Then \((a+4)mod3=(2+4)mod3=6mod3=0mod3\). Hence a+4 is divisible by 3.

Answer 3

\[a \equiv0\mod{3} \implies a \text{ is divisible by 3.} \]\[a \equiv 1 \mod3 \implies a + 2 \text{ is divisible by 3.}\]\[a \equiv 2 \mod 3 \implies a + 4 \text{ is divisible by 3.}\]

Answer 4

Does the above proof I provided work?

Answer 5

Basically just look at what values mod3 a can take, then see what you would have to add (mod3) to get 0mod3.
I think your proof is fine, maybe just say that 4mod3=1mod3, and 2mod3+1mod3=0mod3.

Answer 6

It's a strange question really, it's clear that for any number a, either a, a+1 or a+2 must be divisible by 3. All you have to show really is that (a+4)mod3 =(a+1)mod3.

Answer 7

Yeah, that works as well. :)