<s>[INST] Find the number of solutions to $ \lfloor x \rfloor + \lfloor 2x \rfloor + \lfloor 4x \rfloor + \ldots + \lfloor 32x \rfloor =12345$ 
Find the number of solutions of the equation 
  $$\lfloor x \rfloor + \lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 8x \rfloor + \lfloor 16x \rfloor + \lfloor 32x \rfloor =12345,$$
  where $\lfloor\,\cdot\,\rfloor$ represents the floor function.

My work:
I use the fact that
$$\lfloor nx \rfloor =\sum_{k=0}^{n-1} \left\lfloor x +\frac kn \right\rfloor.$$
So the equation becomes
$$\lfloor x \rfloor +\sum_{k=0}^{1} \left\lfloor x +\frac k2 \right\rfloor +\sum_{k=0}^{3} \left\lfloor x +\frac k4 \right\rfloor +\sum_{k=0}^{7} \left\lfloor x +\frac k8 \right\rfloor   \\
\qquad {}+\sum_{k=0}^{15} \left\lfloor x +\frac k{16} \right\rfloor +\sum_{k=0}^{31} \left\lfloor x+\frac k{32} \right\rfloor   \\
= \lfloor x \rfloor + \left\lfloor x+\frac 12 \right\rfloor + \left\lfloor x+\frac 64 \right\rfloor + \left\lfloor x+\frac{28}{8} \right\rfloor   \\
\qquad {}+ \left\lfloor x+\frac{120}{16} \right\rfloor + \left\lfloor x+\frac{496}{32} \right\rfloor$$
What should I do next?
 [/INST] Your work, while a reasonable approach, does not lead to the answer. A slightly simpler approach does:
Let $p \in \Bbb{Z} = \lfloor x \rfloor$ and let $x= p +q$ with $0\leq q < 1$.
Then the left hand side is
$$L = 63p + \lfloor q\rfloor + \lfloor 2q\rfloor+ \lfloor 4q\rfloor+ \lfloor 8q\rfloor+ \lfloor 16q\rfloor+ \lfloor 32q\rfloor \geq 63p \\
L = 12345 \implies 12345 \geq 63p  \implies p \leq 12345/63 < 196
$$ 
$$L = 63p + \lfloor q\rfloor + \lfloor 2q\rfloor+ \lfloor 4q\rfloor+ \lfloor 8q\rfloor+ \lfloor 16q\rfloor+ \lfloor 32q\rfloor \leq 63p + 1+3+7+15+31 \\
L = 12345 \implies 12345 \leq 63p + 57 \implies p \geq 12288/63 > 195
$$ 
So if a solution exists, then $p$ is an integer between about $195.05$ and $195.95$.  No such integer exists, the the number of solutions is zero.
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