Suppose the set $\mathbb{Z}_{\ge 0}$ of nonnegative integers is partitioned into finitely many arithmetic progressions of the form $a_i \mathbb{Z}_{\ge 0} + b_i$ with $1\leq i\leq n, b_i\ge 0$ and $1\leq a_1 \leq a_2\leq \cdots \leq a_n$. Prove that $a_n = a_{n-1}$.
It could be useful to consider generating functions, but I'm not sure how to come up with the generating functions. We could let the exponents of the functions correspond to elements of the arithmetic sequence $a_i \mathbb{Z}_{\ge 0} + b_i$ for instance. Then we'd get a function like $\prod_{k=0}^\infty (1+x^{a_i k + b_i}$. One way to partition the set of integers is to just choose $n=2, a_1=a_2=2$ and $b_1=0,b_2=1$. So $a_1=a_2$ in that case. I'm not sure if it's useful to consider some sort of contradiction (e.g. if $a_n\neq a_{n-1}$, then integers of a special form may not be in any of the progressions). Maybe some modular arithmetic might be useful since arithmetic progressions are involved (e.g. maybe consider residues modulo the $a_i$'s since all terms in the sequence $a_i \mathbb{Z}_{\ge 0} + b_i$ are congruent to $b_i$ modulo $a_i$)?