# How to solve an equation involving the floor of a number?

## I'm looking for a solving procedure for this type of exercises.

If $[x]$ represents the floor of $x$, solve the equation:

$$\left[\frac{6x+5}8\right]=\frac{15x-7}5$$

a) {$\frac{4}5$}

b) {$\frac{3}4$}

c) {$\frac{7}{15}$,$\frac{4}5$}

d) {$\frac{6}{15}$}

e) {$\frac{1}{2}$,$\frac{3}4$}

f) {$\frac{1}{2}$,$\frac{4}5$}

Can someone please explain how I can solve this equation? Thank you very much!

ps. I asked a similar question, unfortunately, the same method doesn't seem to work.

-
In this case you just substitute advised answers and check if lhs=rhs. In any case, note that lhs is always an integer, so is rhs: $3x - \frac75$ – Ilya May 24 '12 at 14:19
Is there no way to get to the answer without plugging them all in? – Grozav Alex Ioan May 24 '12 at 14:21
@GrozavAlexIoan It is faster to plug in the values than solving the equation. – azarel May 24 '12 at 14:25

Since $\frac{15x-7}{5}$ is required to be an integer, it is of the form $\frac{2+5j}{15}$ for $j$ an arbitrary integer. Thus, the equation to solve becomes $$\left\lfloor\frac{6\big(\frac{2+5j}{15}\big)+5}{8}\right\rfloor = \frac{15\big(\frac{2+5j}{15}\big)-7}{5}\qquad\text{which simplifies to}\quad \left\lfloor\frac{\frac{4}{5}+2j+5}{8}\right\rfloor = j-1$$ Now observe that the function on the right hand side increases quicker than the function on the left hand side. So we can solve the question by finding a maximal solution to the inequality $$j-1\leq\left\lfloor\frac{\frac{4}{5}+2j+5}{8}\right\rfloor$$

A warning: But here we should be careful, because a maximal solution to this inequality is not necessarily a solution to our equality. Also, there might be more solutions than just the maximal solution $j_0$ to this inequality. But since the left hand side and the right hand side are linear in $j$, the set of solutions forms an interval in $\mathbb{Z}$. This means that to give all the solutions, it suffices to find the largest integer below $j_0$ for which the inequality is strict.

Note that for any $n\in\mathbb{Z}$ and $a\in\mathbb{R}$, the inequality $n\leq\lfloor a\rfloor$ holds if and only if $n\leq a$ holds. This helps, since now we can continue to compute $$\begin{split} j-1 \leq \left\lfloor\frac{\frac{4}{5}+2j+5}{8}\right\rfloor & \Leftrightarrow j-1 \leq \frac{\frac{4}{5}+2j+5}{8}\\ & \Leftrightarrow 8j-8\leq \frac{4}{5}+2j+5\\ & \Leftrightarrow 10j\leq 23\\ & \Leftrightarrow j\leq 2. \end{split}$$ Therefore, we get the maximal solution for the equation by substituting $j=2$ in $x=\frac{2+5j}{15}$, i.e. $x=\frac{4}{5}$. This is indeed a solution to the equality. As we have noted before, $\frac{4}{5}$ is not necessarily the only solution. Indeed, it is easy to check that $j=1$ also gives a solution to the equation. But for $j\leq 0$ the inequality is strict, so the only solutions to the equation are $x=\frac{4}{5}$ and $x=\frac{7}{15}$.

Here is a link to the answer to another question, where the OP had to deal with the floor function. Maybe you'll find it useful.

-
Can you please explain how you got to this form? $$\frac{2+5j}{15}+k$$ – Grozav Alex Ioan May 24 '12 at 18:23
@GrozavAlexIoan: For $\frac{15x-7}{5}$ to be an integer, it must be that $15x-7$ is divixible by $5$. Or equivalently, that $15x$ has residue $2$ when dividing by $2$. This means that the fractional part of $x$ must be $\frac{2}{15}$, $\frac{7}{15}$ or $\frac{12}{15}$. I summarized that by saying that $x=\frac{2+5j}{15}+k$ for $j=0,1,2$. – Egbert May 24 '12 at 22:35
Thank you! Very helpful! – Grozav Alex Ioan May 25 '12 at 15:17
I see just now that I could just have written $\frac{2+5j}{15}$ and allowing all the $j$, instead of writing $\frac{2+5j}{15}+k$ and only allowing $j=0,1,2$. I'll rewrite my solution – Egbert May 25 '12 at 16:04
I simplified the proof. If anything is unclear: please ask. – Egbert May 25 '12 at 16:27

If you really want to solve it without using the fact that it's multiple-choice, notice that the fractional part of $\frac{6x+5}{8}$ is $Q=\frac{6x+5}{8}-\frac{15x-7}{5}$. Since $Q$ is a fractional part, $0 \leq Q < 1$; you should be able to work backwards to find bounds on $x$. Once you have those bounds, Ilya's observation that $3x-7/5$ is an integer means there are only finitely many values of $x$ you have to worry about checking, so just try them all (or write a program to do it for you).

-

If $x$ is a solution we have $$0 \le \frac {6x + 5} 8 - \frac {15x - 7} 6 < 1$$ and so $41/90 < x \le 9/10$ and $$87/90 < \frac {6x + 5} 8 \le 13/8$$ Evidently $\lfloor \frac{6 x+ 5} 8 \rfloor$ can only be 0 or 1.

Now you have, only, to solve the equation replacing the lhs with those values.

-

Let $\mathbb{Z} \ni k = \lfloor \frac{6 x+ 5}{8} \rfloor$. Then the equation says $x = \frac{k}{3} + \frac{7}{15}$. Now we write the equation for $k$: $$k = \left\lfloor \frac{k}{4} + \frac{39}{40} \right\rfloor$$ It now remains to test few small values of $k$. Values $k=0$ and $k=1$ work out.

These correspond to $x = \frac{7}{15}$ and $x=\frac{12}{15} = \frac{4}{5}$.

-

When you have to chose between different answers, you don't need to solve the equation, you just need to test the proposed answers.

-