Prove that for all integers $x$ with final digit equal to $5$, $x$ is a multiple of $5$

Question

Is my proof for the statement in question correct?

Let $x \in\mathbb{Z} $ with $x=10r-5$ for some $r \in \mathbb {Z} $. Then for $n \in \mathbb {Z} $ we have $x=5 (2n-1)=5k $ where $k\in\mathbb {Z} $ is by definition odd. Thus, $x$ is divisible by $5$ and hence is a multiple of $5.$

Answer 1

Your proof is correct. Two nitpicks:

• In going from $x = 10r - 5$ to $x = 5(2n-1)$, there's no point in the change of variables, as $n=r$.
• No need for the stipulation $k$ is odd. It's nice that you notice that but it's not relevant to the proof.