Let f be a function defined on all real numbers such that for all x, we have f(x + 5) ≥ f(x) + 5, and f(x + 1) ≤ f(x) + 1. If f(1) = 1, find f(2025).

(In reply to Unsolvable by Jer)

Since f(x) is defined for all real numbers, with some back and forth effort, don't both of your "rule 1" and rule 2" eventually be shown to apply to every integer x, and therefore f(x) must = x?  For example, start with x=0, then work to x=-1, then reexamine x=1 to 6.

Posted by Kenny M on 2025-03-19 06:43:49