Solve $\lfloor{x}\rfloor+\lfloor2x\rfloor+\lfloor4x\rfloor+\lfloor16x\rfloor+\lfloor32x\rfloor=12345$ Solve for $x$
$$\lfloor{x}\rfloor+\lfloor2x\rfloor+\lfloor4x\rfloor+\lfloor16x\rfloor+\lfloor32x\rfloor=12345$$  

I tried to put $x$=$I$+$f$ where $I$ is integer part and $f$ is fractional part but that didn't work.

 A: Let $$f(x)=\lfloor x\rfloor+\lfloor 2x\rfloor+\lfloor 4x\rfloor+\lfloor 16x\rfloor+\lfloor 32x\rfloor\;.$$
By actual calculation $f(224)=55\cdot224=12320$ and $f(224.5)=12347$. Suppose that $224<x<224.5$, say $x=224.5-\epsilon$. Then 
$$\begin{align*}
f(x)&=224+\lfloor 449-2\epsilon\rfloor+\lfloor 898-4\epsilon\rfloor+\lfloor 3592-16\epsilon\rfloor+\lfloor 7184-32\epsilon\rfloor\\
&=(224+449+898+3592+7184)+\lfloor-2\epsilon\rfloor+\lfloor-4\epsilon\rfloor+\lfloor-16\epsilon\rfloor+\lfloor-32\epsilon\rfloor\\
&=12347+\lfloor-2\epsilon\rfloor+\lfloor-4\epsilon\rfloor+\lfloor-16\epsilon\rfloor+\lfloor-32\epsilon\rfloor\\
&\le 12343\;,
\end{align*}$$
since $\lfloor -k\epsilon\rfloor\le-1$ for $k>0$. Thus, the equation $f(x)=12345$ has no solution.
A: As $y-1 < \lfloor y \rfloor \leq y$ you get
$$x+2x+4x+16x+32x -5< \lfloor{x}\rfloor+\lfloor2x\rfloor+\lfloor4x\rfloor+\lfloor16x\rfloor+\lfloor32x\rfloor \\ \leq x+2x+4x+16x +32 x$$
Thus
$$55x-5 < 12345 \leq 55x$$
or
$$11x-1 < 2469 \leq 11x$$
This implies
$$ 224.\bar{45} \leq x \leq 224.\bar{54}$$
Now all you need is to see the intervals for $2x, 4x, 16x, 32x$ and see how many choices you have for each.
