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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?

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  • $\begingroup$ Did you mean [] is the floor function? If so, please write it in the post, not only in a tag. "Greatest integer function" is unclear. $\endgroup$ – user351579 Jul 6 '16 at 15:16
  • $\begingroup$ Yes,I have written greatest integer function in question $\endgroup$ – Aakash Kumar Jul 6 '16 at 15:17
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    $\begingroup$ Do you mean [n] is the greatest integer smaller than or equal to n? You should make that clearer. (in the question, not in a tag) $\endgroup$ – user351579 Jul 6 '16 at 15:19
  • $\begingroup$ It is floor function $\endgroup$ – Aakash Kumar Jul 6 '16 at 15:19
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    $\begingroup$ BTW, the notation $[x]$ for greatest integer not $>x$, is a bit old fashioned. Modern books tend to use the floor function $\lfloor x \rfloor$. $\endgroup$ – Mark Fischler Jul 6 '16 at 15:30
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There are no solutions. For $x\to196^-$ the function value is $12342$, and for $x=196$ it is $12348$, with no values in between.

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  • $\begingroup$ Without putting value there is no way to solve it ? $\endgroup$ – Aakash Kumar Jul 6 '16 at 15:39
  • $\begingroup$ @AakashKumar: Not sure what you mean by "without putting value". I solved the corresponding equation without the floor functions, $63x=12345$, yielding $x=\frac{12345}{63}\approx195.95$, which shows that any solution would have to be between that value and $196$; then it just remains to find the limit as $x\to196^-$, which you get by noting that each floor function drops by $1$ when $x$ drops below an integer. $\endgroup$ – joriki Jul 6 '16 at 15:42
  • $\begingroup$ How it show that x must lies between $x=195.95 and 196$ $\endgroup$ – Aakash Kumar Jul 6 '16 at 16:17
  • $\begingroup$ @AakashKumar: Since the floor function is at most equal to its argument, we need at least $x=\frac{12345}{63}$, since that would solve the equation without any deductions from the floor functions. On the other hand, since for integer $x$ the floor function is equal to its argument, we must have $x\lt 196$, since $x=\frac{12345}{63}$ is already enough if we replace the floor functions by their arguments. $\endgroup$ – joriki Jul 6 '16 at 16:21
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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.

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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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  • $\begingroup$ You probably wanted to say $p\leq \frac{12345}{63}<196$, which I took the liberty to edit, along with two other typos. $\endgroup$ – Batominovski Jul 6 '16 at 15:44
  • $\begingroup$ @Mark Shouldn't it be $12345\ge63p\ge12345-57$? $\endgroup$ – Jack's wasted life Jul 6 '16 at 15:44
  • $\begingroup$ In your step ofsecond left hand side you have taken 195.95<195 $\endgroup$ – Aakash Kumar Jul 6 '16 at 15:54
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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 $f$.

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