Given the inequality: $$\left\lfloor \frac{k_1 - x}{k_2} \right\rfloor \le k_e $$ How can I transform it to find the smallest $x$ which satisfies the relation above? Thanks
2026-04-02 21:38:06.1775165886
Resolution of an inequality with floor
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The inequality is the same as $$\frac{k_1-x}{k_2}<\lfloor k_e\rfloor+1$$