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I got a little trouble solving equations that involve floor function in an efficient way.

For example :

$$ \left\lfloor\frac{x+3}{2}\right\rfloor = \frac{4x+5}{3} $$

In the one above, I get that you basically let $$ \frac{4x+5}{3} = k $$ and then inserting $k$ in the left side, take $k = 8l, 8l+1$, and so on and test it.

If there's a better solution to the one above plese tell me.

My main problem is when it comes down to functions that have multiple floors such as :

$$ \left\lfloor \frac{x+1}{3}\right\rfloor + \left\lfloor\frac{2x+5}{6}\right\rfloor = \frac{3x-5}{2} $$

Using the same method for each of them and then intersecting the solutions should give me the right answer but is there a faster way to solve equations like this ?

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    $\begingroup$ use the fact that : $\lfloor x \rfloor=n \Leftrightarrow n\leq x <n+1$ $\endgroup$
    – L F
    Commented May 10, 2015 at 15:35
  • $\begingroup$ @LuisFelipeVillavicencioLopez tought about that but how do I use that in things that involve fractions? $\endgroup$ Commented May 10, 2015 at 15:36
  • $\begingroup$ $\frac{4x+5}{3}$ must be in $\mathbb{Z}$, you can add $-\frac{4x+5}{3}$ to your inequality and solve the equation for $x$ and then $\cap \mathbb{Z}$ $\endgroup$
    – L F
    Commented May 10, 2015 at 15:42
  • $\begingroup$ @LuisFelipeVillavicencioLopez still don't get it. $\endgroup$ Commented May 10, 2015 at 15:49
  • $\begingroup$ @lab bhattacharjee shows you a better way, i was telling you the first line of him, his second line is better than mine, he has a general method for this $\endgroup$
    – L F
    Commented May 10, 2015 at 15:59

4 Answers 4

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As the left hand side is integer, so should be $\dfrac{3x-5}2\iff2|3(x-1)\iff2|(x-1)\implies x$ is odd (assuming $x$ to be an integer)

Again as lcm$(3,6)$ we need to test for $x\equiv0,1,2,3,4,5\pmod6$

But as $x$ is odd, $x\equiv1,3,5\pmod6$

If $x=6b+1$

$$\left\lfloor\frac{x+1}{3}\right\rfloor + \left\lfloor\frac{2x+5}{6}\right\rfloor=\left\lfloor\frac{6b+1+1}{3}\right\rfloor + \left\lfloor\frac{2(6b+1)+5}{6}\right\rfloor=2b+(2b+1)=4b+1$$

and $$\dfrac{3x-5}2=\dfrac{3(6b+1)-5}2=9b-1$$

and so on

If $x$ is not necessarily an integer, $\dfrac{3x-5}2+I\iff x=\dfrac{5+2I}3$

Check for $I\equiv0,1,2\pmod3$

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Since $\dfrac{4x+5}3 = \left\lfloor \dfrac{x+3}2 \right \rfloor$ is an integer, we need $3$ to divide $4x+5$, i.e., $$\dfrac{4x+5}3 = m \in \mathbb{Z} \implies x = \dfrac{3m-5}4 \text{ where }m \in \mathbb{Z}$$ Hence, $$\left\lfloor \dfrac{x+3}2 \right\rfloor = m \implies \dfrac{x+3}2 = m + e \implies x+3 = 2m+2e \implies x = 2m-3+2e$$ where $e \in[0,1)$. Hence, we need $$\dfrac{3m-5}4 = 2m-3+2e \implies 3m-5 =8m-12+8e \implies 8e = -5m+7$$ This gives us $e = \dfrac{7-5m}8$. Since $e \in [0,1)$, we have $7-5m \in [0,8) \implies m = 0,1$. Hence, $$x=-\dfrac54,-\dfrac12$$

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  • $\begingroup$ If you use that in the initial equation it gives : 2 = 1. Didn't check your math but the main idea is that you also missed a solution, it has 2. $\endgroup$ Commented May 10, 2015 at 15:55
  • $\begingroup$ @user1640736 Corrected. It was due to some sign error. $\endgroup$
    – Adhvaitha
    Commented May 10, 2015 at 15:57
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**Below I have provided a solution $\mathbf{\left(\frac{4x+5}{3}=t\right)}$ $,\,t\in\mathbb z$

$\rightarrow$ $$x=\left(\frac{3t-5}{4}\right)$$ $$\lfloor\left(\frac{\frac{3t-5}{4}+3}{2}\right)\rfloor=t$$ $\rightarrow$ $$\lfloor\left(\frac{3t+7}{8}\right)\rfloor=t$$ $$t\le\left(\frac{3t+7}{8}\right)\lt t+1$$ $$t\le\left(\frac{7}{5}\right) ,t\gt\left(\frac{-1}{5}\right)$$ $\rightarrow$ $$t=0 , 1$$ $$x=\left(\frac{-5}{4}\right) or x=\left(\frac{-1}{2}\right)$$ thank you for your attention

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    $\begingroup$ Formatting tips here. $\endgroup$
    – Em.
    Commented Jan 9, 2016 at 9:00
  • $\begingroup$ Have I got enough? $\endgroup$
    – d.v
    Commented Jan 9, 2016 at 15:09
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$$\lfloor\frac{{x}+\mathrm{3}}{\mathrm{2}}\rfloor=\frac{\mathrm{4}{x}+\mathrm{5}}{\mathrm{3}}\in{Z} \\ $$ $$\frac{{x}+\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{4}{x}+\mathrm{5}}{\mathrm{3}}=\left[\mathrm{0};\mathrm{1}\right) \\ $$ $$\frac{-\mathrm{5}{x}−\mathrm{1}}{\mathrm{6}}=\left[\mathrm{0};\mathrm{1}\right) \\ $$ $${x}=\frac{\mathrm{6}\left[\mathrm{0};\mathrm{1}\right)+\mathrm{1}}{-\mathrm{5}}=\frac{\left(-\mathrm{7};-\mathrm{1}\right]}{\mathrm{5}} \\ $$ $$\frac{\frac{\mathrm{4}}{\mathrm{5}}\left(-\mathrm{7}:-\mathrm{1}\right]+\mathrm{5}}{\mathrm{3}}=\left[\mathrm{0};\mathrm{1}\right] \\ $$ $$\frac{\mathrm{4}{x}+\mathrm{5}}{\mathrm{3}}=\left[\mathrm{0};\mathrm{1}\right] \\ $$ $${x}=\frac{\mathrm{3}\left[\mathrm{0};\mathrm{1}\right]−\mathrm{5}}{\mathrm{4}}=\left\{\frac{-\mathrm{5}}{\mathrm{4}},\frac{-\mathrm{1}}{\mathrm{2}}\right\} \\ $$

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