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# ​ Help me. So '[]' means integer division!!!

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[x/2]+[x/4]+[x/8]+507=x

[x/2],[x/4],[x/8]- INTEGER DIVISIONS!!!

Feb 11, 2019

#1
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$$\frac{x}{2}+\frac{x}{4}+\frac{x}{8}=\frac{7x}{8}$$, and this plus 507 equals x.

$$\frac{7x}{8}+507=x, \frac{x}{8}=507$$

Can you find x from here? Just multiply each side by eight.

You are very welcome!

:P

Feb 11, 2019
#2
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U didn't know what mean integer division like me(

Guest Feb 11, 2019
#3
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I think by "Integer Division", you mean the "Floor Function"!.

Using Numberline graph, it gives the following solutions:

x =4039, 4043, 4045, 4046, 4049, 4050, 4052 and 4056.

Feb 11, 2019
#5
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Yes, thx. I just from Russia and i didn't know how it translete right on Eng))) I just gave you the literal translation))

Guest Feb 12, 2019
#6
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[x/2]+[x/4]+[x/8]+507=x

[x/2],[x/4],[x/8]- INTEGER DIVISIONS!!!

I assume the "Floor Function": $$\left\lfloor\dfrac{x}{2}\right\rfloor+\left\lfloor\dfrac{x}{4}\right\rfloor+\left\lfloor\dfrac{x}{8}\right\rfloor+507=x$$

So x is an integer.

We rearrange:

$$\left\lfloor\dfrac{4x}{8}\right\rfloor+\left\lfloor\dfrac{2x}{8}\right\rfloor+\left\lfloor\dfrac{x}{8}\right\rfloor+507=x$$

We substitute: $$y= \dfrac{x}{8}$$
$$\left\lfloor 4y \right\rfloor+\left\lfloor 2y \right\rfloor+\left\lfloor y \right\rfloor+507=x$$

$$\text{The fractional part of \mathbf{y} is \alpha and 0 < \alpha < 1 } \\ \text{The integer part of \mathbf{y} is n }\\ \text{So \mathbf{y = n+\alpha}} \\ \text{\mathbf{x = \lfloor 8y \rfloor} = \lfloor 8(n+\alpha)\rfloor=8n+\lfloor 8\alpha\rfloor}$$

$$\begin{array}{|rcll|} \hline \left\lfloor 4(n+\alpha)\right\rfloor+\left\lfloor 2(n+\alpha) \right\rfloor+\left\lfloor n+\alpha\right\rfloor+507 &=& 8n+\lfloor 8\alpha\rfloor \\\\ \mathbf{\left\lfloor 4n+4\alpha\right\rfloor+\left\lfloor 2n+2\alpha \right\rfloor+\left\lfloor n+\alpha \right\rfloor+507}&\mathbf{=}& \mathbf{8n+\lfloor 8\alpha\rfloor} \\ \hline \end{array}$$

We divide alpha into 8 parts:

$$0 < \alpha < \frac{1}{8},\ \frac{1}{8} \le \alpha < \frac{2}{8},\ \frac{2}{8} \le \alpha < \frac{3}{8},\ \frac{3}{8} \le \alpha < \frac{4}{8} ,\ \frac{4}{8} \le \alpha < \frac{5}{8},\ \frac{5}{8} \le \alpha < \frac{6}{8},\ \frac{6}{8} \le \alpha < \frac{7}{8},\ \frac{7}{8} \le \alpha < \frac{8}{8}$$

$$\begin{array}{|r|r|rcl|} \hline \text{part} & \text{domain} & \\ \hline 1. & 0 < \alpha < \frac{1}{8} & \left\lfloor 4n+4\alpha\right\rfloor+\left\lfloor 2n+2\alpha \right\rfloor+\left\lfloor n+\alpha \right\rfloor+507& =& 8n+\lfloor 8\alpha\rfloor \\ && 4n +2n+n+507 &=& 8n+0 \\ && n &=& 507 \\ && x &=& 8n + \lfloor 8\alpha\rfloor \\ && x &=& 8\cdot 507 \\ && \mathbf{x} & \mathbf{=} & \mathbf{4056} \\ \hline 2. & \frac{1}{8} \le \alpha < \frac{2}{8} & \left\lfloor 4n+4\alpha\right\rfloor+\left\lfloor 2n+2\alpha \right\rfloor+\left\lfloor n+\alpha \right\rfloor+507& =& 8n+\lfloor 8\alpha\rfloor \\ && 4n +2n+n+507 &=& 8n+1 \\ && n &=& 506 \\ && x &=& 8n + \lfloor 8\alpha\rfloor \\ && x &=& 8\cdot 506+1 \\ && \mathbf{x} & \mathbf{=} & \mathbf{4049} \\ \hline 3. & \frac{2}{8} \le \alpha < \frac{3}{8} & \left\lfloor 4n+4\alpha\right\rfloor+\left\lfloor 2n+2\alpha \right\rfloor+\left\lfloor n+\alpha \right\rfloor+507& =& 8n+\lfloor 8\alpha\rfloor \\ && 4n +1+2n+n+507 &=& 8n+2 \\ && n &=& 506 \\ && x &=& 8n + \lfloor 8\alpha\rfloor \\ && x &=& 8\cdot 506+2 \\ && \mathbf{x} & \mathbf{=} & \mathbf{4050} \\ \hline 4. & \frac{3}{8} \le \alpha < \frac{4}{8} & \left\lfloor 4n+4\alpha\right\rfloor+\left\lfloor 2n+2\alpha \right\rfloor+\left\lfloor n+\alpha \right\rfloor+507& =& 8n+\lfloor 8\alpha\rfloor \\ && 4n +1+2n+n+507 &=& 8n+3 \\ && n &=& 505 \\ && x &=& 8n + \lfloor 8\alpha\rfloor \\ && x &=& 8\cdot 505+3 \\ && \mathbf{x} & \mathbf{=} & \mathbf{4043} \\ \hline 5. & \frac{4}{8} \le \alpha < \frac{5}{8} & \left\lfloor 4n+4\alpha\right\rfloor+\left\lfloor 2n+2\alpha \right\rfloor+\left\lfloor n+\alpha \right\rfloor+507& =& 8n+\lfloor 8\alpha\rfloor \\ && 4n +2+2n+1+n+507 &=& 8n+4 \\ && n &=& 506 \\ && x &=& 8n + \lfloor 8\alpha\rfloor \\ && x &=& 8\cdot 506+4 \\ && \mathbf{x} & \mathbf{=} & \mathbf{4052} \\ \hline 6. & \frac{5}{8} \le \alpha < \frac{6}{8} & \left\lfloor 4n+4\alpha\right\rfloor+\left\lfloor 2n+2\alpha \right\rfloor+\left\lfloor n+\alpha \right\rfloor+507& =& 8n+\lfloor 8\alpha\rfloor \\ && 4n +2+2n+1+n+507 &=& 8n+5 \\ && n &=& 505 \\ && x &=& 8n + \lfloor 8\alpha\rfloor \\ && x &=& 8\cdot 505+5 \\ && \mathbf{x} & \mathbf{=} & \mathbf{4045} \\ \hline 7. & \frac{6}{8} \le \alpha < \frac{7}{8} & \left\lfloor 4n+4\alpha\right\rfloor+\left\lfloor 2n+2\alpha \right\rfloor+\left\lfloor n+\alpha \right\rfloor+507& =& 8n+\lfloor 8\alpha\rfloor \\ && 4n +3+2n+1+n+507 &=& 8n+6 \\ && n &=& 505 \\ && x &=& 8n + \lfloor 8\alpha\rfloor \\ && x &=& 8\cdot 505+6 \\ && \mathbf{x} & \mathbf{=} & \mathbf{4046} \\ \hline 8. & \frac{7}{8} \le \alpha < \frac{8}{8} & \left\lfloor 4n+4\alpha\right\rfloor+\left\lfloor 2n+2\alpha \right\rfloor+\left\lfloor n+\alpha \right\rfloor+507& =& 8n+\lfloor 8\alpha\rfloor \\ && 4n +3+2n+1+n+507 &=& 8n+7 \\ && n &=& 504 \\ && x &=& 8n + \lfloor 8\alpha\rfloor \\ && x &=& 8\cdot 504+7 \\ && \mathbf{x} & \mathbf{=} & \mathbf{4039} \\ \hline \end{array}$$

Feb 12, 2019
edited by heureka  Feb 13, 2019