Solve the equation: $\lfloor{x}\rfloor+\lfloor{2x}\rfloor=\frac{5x-1}{2}$

Question

Solve the equation:

$$\lfloor{x}\rfloor+\lfloor{2x}\rfloor=\frac{5x-1}{2}$$

Unfortunately I don't know how to approach this problem. I tried all the properties and I even tried Hermite's identity, but no luck.

Answer 1

Hint: The LHS is an integer, so the RHS must be one as well. See if you can use this facts to get rid of the floor function and replace it with a linear polynomial of $x$

Answer 2

The easiest thing to note is that $[x] + [2x]$ is an integer.

So $\frac {5x -1}2$ is an integer.

Let $\frac {5x -1}2 = W\in \mathbb Z$ then

$x =\frac {2W+1}5$ so $x$ is a fraction of an odd number numerator over denominator $5$.

So let $x = n + \frac k5$ where $n,k \in \mathbb Z$ and $0 \le k < 5$. (In other words where $2W+1 = 5n+k$)

Then $[x] = n$ and $2x = 2n + \frac {2k}5$ and $[2x] = 2n$ if $\frac {2k}5 < 1$ or $[2x] =2n + 1$ of $1 \le \frac {2k}5 < 2$

So

Case 1: $\frac {2k}5 < 1$ or $2k < 5$ or $k =0,1,2$.

Then $[x] + [2x] = n + 2n = 3n$ and $\frac {5(n+\frac k5)-1}2 = \frac {5n+k-1}2=3n$

$5n +k -1 = 6n$

$n = k-1$ so $n=-1;k=0;x=-1$ or $n=0;k=1;x=\frac 15$ or $n=1;k=2;x=1+\frac 25$.

Case 2: $1\le \frac {2k}5 < 2$ or $5\le 2k < 10$ or $k=3,4$.

Then $[x] +[2x] = n +2n+1 = 3n+1$ and $\frac {5(n+\frac k5)-1}2=\frac {5n+k-1}2 = 3n+1$

$5n+k-1 = 6n +2$

$n = k-3$

And so $n=0;k=3;x=\frac 35$ or $n=1;k=4; x=\frac 95$.

Answer 3

Method I: Use inequality $x-1 < [x] \leq x$

$[x]+[2x] \leq x+2x = 3x \Rightarrow \frac{5x-1}{2} \leq 3x \Rightarrow -1 \leq x$

$x-1+2x-1 < [x]+[2x] \Rightarrow 3x-2 <\frac{5x-1}{2} \Rightarrow 6x-4 <5x-1 \Rightarrow x<3$

you can find a range for x: $-1 \leq x < 3$

Then you can divide it into 8 ranges in worst cases and solve them one by one. e.g. $-1 \leq x<-0.5, -0.5 \leq x <0$, etc (Deal to the term $[2x]$, I divide the ranges into steps of $0.5$)

We have $\frac{5x-1}{2}=-3,-2,0,?,?,?,?,?$ for those 8 ranges.

Then the corresponding solutions are $?,?,?,?,?,?,?,?$

Check with their ranges. Some of them are rejected, $?,?,?,?,?$ remained. They are the solutions. I keep some number as ? to let you try.

Method II: Use @user2661923 's method. You do not even need inequality.

Case I: Let $x = P+r , 0 \leq r <0.5$, P is integer.

$P+2P = \frac{5(P+r)-1}{2} \Rightarrow P=5r-1$

Put in $r=0,0.2,0.4$ give you 3 solutions.

Case II: Let $x = P+r , 0.5 \leq r <1$, P is integer.

$P+2P+1 = \frac{5(P+r)-1}{2} \Rightarrow P=5r-3$

Put in $r=0.6,0.8$ give you 2 solutions.