RMO 2015

Normally I use $\{x\}$ and $\lfloor x \rfloor$ to represent fractional and integral part of $x$ , respectively.

The trick of this question is $4<a<5$ , so we know for sure that $\lfloor a \rfloor=4$ . Furthermore, we have an identity $\{a\}=a-\lfloor a \rfloor$ , thus we get $\{a\}=a-4$ . The expression is transformed into $a(12-2a).$ Suppose that $a(12-2a)=k$ , where $k$ is an integer as given. We have $2a^2-12a+k=0$ , whose solutions are of the form $a=\frac{6 \pm \sqrt{36-2k}}{2}.$ Apply this form to the inequality $4<a<5$ , we have $4< \frac{6+\sqrt{36-2k}}{2} <5,$ $2< \sqrt{36-2k} <4,$ $4< 36-2k <16,$ $10< k <16.$ Consequently, there are 5 known possible values of $a$ that satisfy our condition, which occur when $k=11,12,13,14,15$ . These values of $a$ are real, since $k<18$ implies $36-2k>0$ .

On the other hand, if we start the inequality by $\frac{6-\sqrt{36-2k}}{2}$ , this is impossible because it leads to $2<-\sqrt{36-2k}<4$ .

Therefore, there are exactly 5 possible values of $a$ .