Find the number of solutions to $ \lfloor x \rfloor + \lfloor 2x \rfloor + \lfloor 4x \rfloor + \ldots + \lfloor 32x \rfloor =12345$

Question

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?

Answer 1

There are no solutions. For $x\to196^-$ the function value is $12342$, and for $x=196$ it is $12348$, with no values in between.

Answer 2

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.

Answer 3

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.

Answer 4

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 $f$.

Answer 5

$$\underset{{n}=\mathrm{0}} {\overset{\mathrm{5}} {\sum}}\lfloor\mathrm{2}^{{n}} {x}\rfloor=\mathrm{12345} \\ $$ $${x}\underset{{n}=\mathrm{0}} {\overset{\mathrm{5}} {\sum}}\left(\mathrm{2}^{{n}} −{f}_{{n}} \right)=\mathrm{12345} \\ $$ $$\mathrm{63}{x}=\left[\mathrm{12345};\mathrm{12351}\right) \\ $$ $$\lfloor{x}\rfloor=\mathrm{196} \\ $$ $$\lfloor\mathrm{2}{x}\rfloor=\mathrm{392} \\ $$ $$\lfloor\mathrm{4}{x}\rfloor=\mathrm{784} \\ $$ $$\lfloor\mathrm{8}{x}\rfloor=\mathrm{1568} \\ $$ $$\lfloor\mathrm{16}{x}\rfloor=\mathrm{3136} \\ $$ $$\lfloor\mathrm{32}{x}\rfloor=\left[\mathrm{6271};\mathrm{6273}\right] \\ $$ $$\lfloor\mathrm{32}{x}\rfloor=\underset{{n}=\mathrm{0}} {\overset{\mathrm{5}} {\sum}}\lfloor\mathrm{2}^{{n}} {x}\rfloor−\underset{{n}=\mathrm{0}} {\overset{\mathrm{4}} {\sum}}\lfloor\mathrm{2}^{{n}} {x}\rfloor \\ $$ $$\lfloor\mathrm{32}{x}\rfloor=\mathrm{6269} \\ $$ $$\left[{NO}\:{SOLUTION}\right] \\ $$