jmc

Since $0\le \{x\}<1,$ we have $0\le 100\{x\}<100,$ so $5\le 5+100\{x\}<105.$ Thus, $$5\le \lfloor x\rfloor <105.$$Since $\lfloor x\rfloor$ is an integer, the possible values of $\lfloor x\rfloor$ are $5,6,\dots ,104.$ For each of these values of $\lfloor x\rfloor ,$ we get a corresponding value $$\{x\}=\frac{\lfloor x\rfloor -5}{100}=0.01\lfloor x\rfloor -0.05,$$and then we have $$x=\lfloor x\rfloor +\{x\}=1.01\lfloor x\rfloor -\mathrm{0.05.}$$To maximize $x,$ we choose $\lfloor x\rfloor =104,$ giving $$x=1.01\cdot 104-0.05=\boxed{104.99}.$$