From Bóna's [A Walk through Combinatorics](https://books.google.com/books?id=x9tqMFkkTqQC):



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> Prove or disprove that if we partition the positive integers into finitely many arithmetic progressions then there will be at least one arithmetic progression whose difference and initial term are equal.
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A variant of this problems asks to prove that there will be at least one arithmetic progression whose difference divides the initial term, which has an elementary proof. I suspect the same arithmetic progression will have the initial term equal to the difference. However, I cannot prove it.



Partitioning the positive integers into finitely many arithmetic progressions
Any progression (if they are all infinite, of course, but otherwise the statement is clearly wrong) should have its initial term $a$ not greater than the difference $d$. Indeed, if $a>d$, and $a-d$ is covered by another progression $P$ with difference $d\_1$, then $a-d+dd\_1$ is covered twice --- a contradiction. Therefore if $a$ is divisible by $d$, we must have $a=d$. And such $a$ exists as you noted in OP (a short proof: take a progression which contains the product of all differences).


