$x\geq0$
in terms of $k$
that satisfy $\lfloor kx \rfloor = (k+1)\lfloor x \rfloor,$
where $k>0$
is an integer and $\lfloor r \rfloor$
is the greatest integer less than or equal to $r.$
You may wish to consider $k=2$
first.

 Let
$f(x)=x^3  3x^2  3x + 1$
and$g(x) = x+1.$
For which values of$x$
is$g(x) > f(x)?$
 Find the set of solutions for
$x$
that satisfies$\lfloor \frac{x}{2} \rfloor = 4.$
 Let

By rewriting
$x,$
how can one simplify the equation? 
… in particular, let
$x = n+f,$
where$n$
is the integer part and$f$
is the fractional part$(0\le f<1).$

Consider the relation between
$\lfloor n+f \rfloor$
and$n.$

After simplifying, try considering the different possible cases for
$n.$

… and the corresponding values of
$f?$

In this solution we use the notation
$x\in[a,b)$
to mean$a \leq x < b.$
For the general case, let
$x=n+f$
where$n$
is the integer part and$f\in[0,1)$
is the fractional part. We have$\lfloor kn+kf \rfloor=(k+1)\lfloor n+f \rfloor$
which gives$kn+\lfloor kf \rfloor=(k+1)n$
and finally$\lfloor kf \rfloor=n.$
This means that$n$
can only take the values$n=0,1,\ldots,k1.$
Taking these in turn we get: For
$n=0,$
$f\in[0,\frac{1}{k})$
and so$x\in[0,\frac{1}{k}).$
 For
$n=1,$
$f\in[\frac{1}{k},\frac{2}{k})$
and so$x\in[1+\frac{1}{k},1+\frac{2}{k}).$
$\cdots$
 For
$n=r,$
$f\in[\frac{r}{k},\frac{r+1}{k})$
and so$x\in[r+\frac{r}{k},r+\frac{r+1}{k}).$
$\cdots$
In the end, we take the union of all these sets up to
$n=k1,$
which is formally written as$x\in\bigcup\limits_{n=0}^{k1} \big[n+\frac{n}{k}, n+\frac{n+1}{k}\big).$
Note: this notation is not required for full credit.
 For