An integer $n > 7$ has form $r + 3k,\ r \in \{8,9,10\}\ $ [remainder systems] I want to prove the following statement, related to a coins problem.
All integers greater than 7 can be expressed as 8 + 3k or 9 + 3k or 10 + 3k for some non-negative integer k.
I believe it can be done by strong induction, but I'm not quite ready to tackle that yet and am looking for a more elementary proof. I'm not sure how to approach it. It is obvious from investigating the integer number line, but I want something more formal.
$(\forall n \in \mathbb{N}), n > 7 \implies (n = 8 + 3k  \lor n = 9 + 3k \lor n = 10 +3k) $ for some $k \in Z_{\ge 0}$
Did I get the statement of the proposition right?
What would be a good way to proceed? Contradiction? Cases?

You got the statement right. Now, let $r$ be the remainder of the division of $n$ by $3$. Then $n=3k+r$ for some $k\in\mathbb Z_+$ and $r\in\{0,1,2\}$. Actually, since $n>7$, $k$ is at least $2$, with $k\geqslant3$ if $n\geqslant9$. But then:


*

*if $r=2$, $n=3(k-2)+6+2=3(k-2)+8$;

*otherwise, $n=3(k-3)+9+r$, and $9+r$ is either $9$ or $10$.

