Prove that for all integers $n$, exactly one of $n$, $2n − 1$ and $2n + 1$ is divisible by $3$.
2026-04-13 01:37:05.1776044225
Modular Arithmetic Divisibility
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Hint: By the division algorithm you can write $n$ as a multiple of $3$ with some remainder $r$. ie. there exists $d\in \mathbb Z$ such that $$n=d*3+r$$ where $r\in \{0,1,2\}$. Compare $r$ for $2n-1,n,2n+1$.