Enough to check $x<3*5*7*11$ $165=3*5*11$ and so $7|x$ $\{\frac{x}{3}\}=\frac{a}{3}, \{\frac{x}{5}\}=\frac{b}{5},\{\frac{x}{11}\}=\frac{c}{11}$ $\frac{a}{3}+\frac{b}{5}+\frac{c}{11}=\frac{248}{165}$ $55a+33b+15c=248$ and $a \leq 2, b \leq 4,c \leq 10$ $4c \equiv 6 \pmod{11} \to c =7$ $3b \equiv 3 \pmod {5} \to b=1$ $a \equiv 2 \pmod {3} \to a=2$ So $x \equiv 2 \pmod {3}, x \equiv 1 \pmod {5}, x \equiv 0 \pmod {7}, x \equiv 7 \pmod {11}$ and so $x \equiv 161 \pmod{3*5*7*11}$ And so $\{\frac{2x}{3}\}+\{\frac{2x}{5}\}+\{\frac{2x}{7}\}+\{\frac{2x}{11}\}=\frac{2}{3}+\frac{2}{5}+\frac{5}{11} = \frac{163}{165}$

Should it not be $\frac{1}{3}+\frac{2}{5}+\frac{3}{11}=\frac{166}{165}$ in the last line?