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?
 A: There are no solutions. For $x\to196^-$ the function value is $12342$, and for $x=196$ it is $12348$, with no values in between.
A: 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.
A: Let 
$$
f(x) = \lfloor x \rfloor + \lfloor 2x \rfloor + \lfloor 4x \rfloor
+ \lfloor 8x \rfloor + \lfloor 16x \rfloor + \lfloor 32x \rfloor.
$$
For integers $n$ we have $f(n)=63n$, but for a small number to the left of $n$ all terms $\lfloor x \rfloor, \lfloor 2x \rfloor,\dots$ decrease by one. So at every integer $n$ we have that $f$ jumps from $63n-6$ to $63n$. By calculation we have $$f(196)=12348,$$ so at $n=196$ we find that $f$ jumps from $12342$ to $12348$. Since $f$ is increasing we find that $f(x)=12345$ has no solutions. 
A: Alt. solution: Write the fractional part of $x$ in binary:
$$
x = n + 0.b_1 b_2 b_3 \ldots = n + \sum_{i=1}^\infty \frac{b_i}{2^i},
$$
where $n \in \mathbb{Z}$ and $b_i \in \{0,1\}$.
Also give your function $\mathbb{R} \to \mathbb{R}$ a name:
$$
f(x) = \lfloor x \rfloor + \lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 8x \rfloor + \lfloor 16x \rfloor + \lfloor 32x \rfloor 
$$
Then
\begin{align*}
f(x)
&= \lfloor x \rfloor + \lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 8x \rfloor + \lfloor 16x \rfloor + \lfloor 32x \rfloor \\
&= n + (2n + b_1) + (4n + 2b_1 + b_2) + \cdots + (32n + 16b_1 + 8b_2 + 4b_3 + 2b_4 + b_5) \\
&= 63n + 31b_1 + 15b_2 + 7b_3 + 3b_4 + b_5.
\end{align*}
We can choose $b_i \in \{0,1\}$ arbitrarily,
so it follows that the image of $f$ is exactly
the set of integers whose remainder is between $0$ and $31 + 15 + 7 + 3 + 1 = 57$, mod $63$.
In particular, $12345 \equiv 60 \pmod{63}$, so $12345$ is not in the image of $f$.
