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floor of (an+b)/(cn+d) is surjective

Source: Romanian NMO 2021 grade 10 P2

April 15, 2023
floor functionalgebrafunction

Problem Statement

Let a,b,c,d∈Z≥0a,b,c,d\in\mathbb{Z}_{\ge 0}, d≠0d\ne 0 and the function f:Z≥0→Z≥0f:\mathbb{Z}_{\ge 0}\to\mathbb Z_{\ge 0} defined by f(n)=⌊an+bcn+d⌋ for all n∈Z≥0.f(n)=\left\lfloor \frac{an+b}{cn+d}\right\rfloor\text{ for all } n\in\mathbb{Z}_{\ge 0}. Prove that the following are equivalent:
[*] ff is surjective; [*] c=0c=0, b<db<d and 0<a≤d0<a\le d.
Tiberiu Trif
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