Selection tests for the Balkan Mathematical Olympiad 2013

Notice first that if $\lfloor x\rfloor \le 0$ then $$\lfloor x^2\rfloor -10\lfloor x\rfloor +24\ge 24>0$$ This means that if $x$ is a solution to the equation then $\lfloor x\rfloor \ge 1$. Let $x$ be a solution to the equation, $m=\lfloor x\rfloor \ge 1$, and $r=x-\lfloor x\rfloor \ge 0$. We have $$0=\lfloor x^2\rfloor -10\lfloor x\rfloor +24=\lfloor r(r+2m)\rfloor +(m-4)(m-6).$$ Because $r(r+2m)\ge 0$, we deduce that $(m-4)(m-6)\le 0$, which is equivalent to $m=4,5$, or $6$.

1. If $m=4$, the equation becomes $\lfloor r(r+8)\rfloor =0$. This is equivalent to $(r+4{)}^2-17<0$ and $(r+4{)}^2-16\ge 0$, and its solutions are $0\le r<\sqrt{17}-4$. This means that the solutions to the equation in this case are $4\le x<\sqrt{17}$.

2. If $m=5$, the equation becomes $\lfloor r(r+10)\rfloor =1$. This is equivalent to $(r+5{)}^2-27<0$ and $(r+5{)}^2-26\ge 0$, and its solutions are $\sqrt{26}-5\le r<\sqrt{27}-5$. This means that the solutions to the equation in this case are $\sqrt{26}\le x<\sqrt{27}$.

3. If $m=6$, the equation becomes $\lfloor r(r+12)\rfloor =0$. This is equivalent to $(r+6{)}^2-37<0$ and $(r+6{)}^2-36\ge 0$, and its solutions are $0\le r<\sqrt{37}-6$. This means that the solutions to the equation in this case are $6\le x<\sqrt{37}$.

Thus, the set of solutions to this equation is $$[4,\sqrt{17})\cup [\sqrt{26},\sqrt{27})\cup [6,\sqrt{37}).$$