An algebra problem by Ilham Saiful Fauzi

$2015x = 2016 \lfloor x \rfloor$

For a value of $\lfloor x \rfloor = a$ , where $a \in Z$ there must be a solution for $x$ .

First, let find the smallest value for $\lfloor x \rfloor$ .

• For $\lfloor x \rfloor = 0$ : $2015x = 0 \rightarrow x=0$ .

Since $\lfloor 0 \rfloor= 0$ , $\lfloor x \rfloor=0$ is a solution.

• For $\lfloor x \rfloor = -1$ : $2015x = -2016 \rightarrow x=-1.000496$ .

Since $\lfloor-1.000496 \rfloor \neq -1$ , $\lfloor x \rfloor= -1$ is not a solution.

So, the smallest value for $\lfloor x \rfloor$ is $0$ .

Next, lets find the largest value for $\lfloor x \rfloor$ .

• For $\lfloor x \rfloor= 1$ : $2015x = 2016 \rightarrow x=1.000496$ .

Since $\lfloor 1.000496 \rfloor = 1$ , $\lfloor x \rfloor= 1$ is a solution.

• For $\lfloor x \rfloor= 2$ : $2015x = 4032 \rightarrow x=2.000993$ .

Since $\lfloor 2.000993 \rfloor= 2$ , $\lfloor x \rfloor = 2$ is a solution.

• For $\lfloor x \rfloor= 3$ : $2015x = 6048 \rightarrow x=3.001489$ .

Since $\lfloor 3.001489 \rfloor = 3$ , $\lfloor x \rfloor = 3$ is a solution.

$\quad \vdots$

• For $\lfloor x \rfloor = 2013$ : $2015x = 2016 \times 2013 = 4058208 \rightarrow x=2013.999007$ .

Since $\lfloor 2013.999007 \rfloor = 2013$ , $\lfloor x \rfloor= 2013$ is a solution.

• For $\lfloor x \rfloor= 2014$ : $2015x = 2016 \times 2014 = 4060224 \rightarrow x=2014.999504$ .

Since $\lfloor 2014.999504 = 2014$ , $\lfloor x \rfloor = 2014$ is a solution.

• For $\lfloor x \rfloor = 2015$ : $2015x = 2016 \times 2015 \rightarrow x=2016$ .

Since $\lfloor 2016 \rfloor\neq 2015$ , $\lfloor x \rfloor= 2015$ is not a solution.

So, the largest value for $\lfloor x \rfloor$ is $2014$ .

The value of $\lfloor x \rfloor$ so that $2015x = 2016 \lfloor x \rfloor$ are $0,1,2,3,\ldots,2013,2014$ .

Since for every value of $\lfloor x \rfloor$ there must be a solution for $x$ , the number of solution of $x$ is $\boxed{2015}$ .

$\begin{aligned} 2015x & = 2016 \lfloor x \rfloor \quad \quad \small \color{#3D99F6}{\text{Since RHS is an integer, LHS must be an integer.} \implies x = \frac m{2015}} \\ 2015 \cdot \color{#3D99F6}{\frac m{2015}} & = 2016 \lfloor x \rfloor \\ \implies \color{#3D99F6}{m} & = 2016 \lfloor x \rfloor \quad \quad \small \color{#3D99F6}{\implies m \text{ is a multiple of 2016, that is } m = 2016n} \end{aligned}$

$\begin{aligned} \implies x & = \frac {2016n}{2015} = n + \frac n{2015} \end{aligned}$

We note that for $2015x = 2016 \lfloor x \rfloor$ to be true $\lfloor x \rfloor$ must be equal to $n$ . Since for $x < 0$ , $\lfloor x \rfloor = n-1$ , $x$ must be $\ge 0$ . For $x \ge 0$ , we note that $\lfloor x \rfloor = n$ if $\dfrac n{2015} < 1\implies n < 2015$ . Therefore, $n=0,1,2,...2014$ and the number of solutions $x$ is $\boxed{2015}$ .

$\begin{aligned} 2015x & =2016 \lfloor{x}\rfloor \\ 2015( \lfloor{x}\rfloor +\{x\}) & =2016 \lfloor{x}\rfloor \\ 2015 \lfloor{x}\rfloor +2015\{x\} & =2016 \lfloor{x}\rfloor \\ 2015\{x\} & = \lfloor{x}\rfloor \\ \{x\} & =\frac{\lfloor{x}\rfloor}{2015} \end{aligned}$

Since $\{x\}$ is a fractional part, then

$0 \leq \{x\} < 1$

$0 \leq \large\frac{\lfloor{x}\rfloor}{2015} < 1$

$0 \leq \lfloor{x}\rfloor < 2015 \implies \lfloor{x}\rfloor = 0,1,2,...,2014$ .

Since there are $2015$ solution for $\lfloor{x}\rfloor$ , then the number of real solution $x$ satisfying $2015x =2016 \lfloor{x}\rfloor$ is $\boxed{2015}$ .