Floors and Fraction

We rewrite the above statement as

$\begin{aligned} x - \lfloor x \rfloor &= 2017 \left ( \dfrac {1}{\lfloor x \rfloor} - \dfrac {1}{x} \right )\\ &= \dfrac {2017(x - \lfloor x \rfloor)}{x\lfloor x \rfloor}\\ \end{aligned}$

Since $x$ is not an integer, it follows $x \neq \lfloor x \rfloor$ , so we can divide by $x - \lfloor x \rfloor$ and rearrange to get

$x \lfloor x \rfloor = 2017$

Case 1: $\lfloor x \rfloor \geq 45$

Then $x >45$ , so $x \lfloor x \rfloor > 45^2 = 2025 > 2017$ .

Case 2: $-44 \leq \lfloor x \rfloor \leq 44$

Then $-44 < x < 45$ , so $x \lfloor x \rfloor < 44 \times 45 = 1980 < 2017$ .

Case 3: $\lfloor x \rfloor \leq -46$

Then $x < -45$ , so $x \lfloor x \rfloor > 45 \times 46 = 2070 > 2017$

Case 4: $\lfloor x \rfloor = -45$

From this, we derive $x = \frac{-2017}{45} = -44 \frac{37}{45}$

Checking in the original expression, we have

$\begin{aligned} LHS &= x + \dfrac {2017}{x} = \dfrac {-2017}{45} + \dfrac{2017}{\frac{-2017}{45}} = -45 - \dfrac {2017}{45}\\ RHS &= \lfloor x \rfloor + \dfrac{2017}{\lfloor x \rfloor} = -45 - \dfrac {2017}{45}\\ \implies LHS &= RHS\\ \end{aligned}$

Therefore, the only solution is $x = \dfrac {-2017}{45}$ , so the answer is $-2017+45=-1972$ .

${ eq }^{ n }\quad can\quad be\quad written\quad as,\\ \\ x-\left\lfloor x \right\rfloor =\cfrac { 2017 }{ x\left\lfloor x \right\rfloor } \left( x-\left\lfloor x \right\rfloor \right) \\ \\ for\quad non\quad integral\quad { sol }^{ n },\\ \\ \cfrac { 2017 }{ x\left\lfloor x \right\rfloor } =1\quad \Rightarrow \quad x\left\lfloor x \right\rfloor =2017\quad \Rightarrow \quad x=\cfrac { 2017 }{ \left\lfloor x \right\rfloor } \\ \\ \left\lfloor x \right\rfloor \quad will\quad be\quad close\quad to\quad \pm \sqrt { 2017 } =\pm 44.9\\ \\ max\quad difference\quad b/w\quad x\quad \& \quad \left\lfloor x \right\rfloor =1\\ \\ Now,hit\quad \& \quad trial\\ \\ \cfrac { 2017 }{ \pm 44 } =\pm 45.84....\quad diff>1\\ \\ \cfrac { 2017 }{ \pm 45 } =\pm 44.82...\quad diff<1\\ \\ \cfrac { 2017 }{ \pm 46 } =\pm 43.84...\quad diff>1\\ \\ only\quad possibility\quad is\quad \left\lfloor x \right\rfloor =-45\\ \\ \Rightarrow x=\cfrac { 2017 }{ -45 } =\cfrac { -2017 }{ 45 } \quad \Rightarrow a+b=-2017+45=-1972\\ \\ \\ \\$

$x - \lfloor x\rfloor = 2017\cdot \frac{x - \lfloor x\rfloor}{x\lfloor x\rfloor}$ .

Since we are searching for non - integral solutions , therefore , $x - \lfloor x\rfloor \neq 0$ .

$\Rightarrow x\lfloor x\rfloor = 2017$ .

$\underline{CASE 1:} \Rightarrow \lfloor x\rfloor > 0$ .

Now , we know that , $\lfloor x\rfloor < x < \lfloor x\rfloor + 1$ .

$\Rightarrow {\lfloor x\rfloor}^2 < x\lfloor x\rfloor = 2017 < {\lfloor x\rfloor}^2 + \lfloor x\rfloor$

This leads to no solution because the first inequality leads to $\lfloor x\rfloor \leq 44$ whereas the second one leads to $\lfloor x\rfloor \geq 45$ .

$\underline{CASE 2 :} \Rightarrow \lfloor x\rfloor < 0$

Now , we know that $\lfloor x\rfloor < x < \lfloor x\rfloor + 1$ .

$\Rightarrow {\lfloor x\rfloor}^2 > x\lfloor x\rfloor = 2017 > {\lfloor x\rfloor}^2 + \lfloor x\rfloor$ .

The first inequality leads to $\lfloor x\rfloor \leq {-45}$ and the second one leads to $\lfloor x\rfloor \geq {-45}$

$\Rightarrow \lfloor x\rfloor = -45$

This leads to $x = \frac{-2017}{45}$ .

Hence , our answer is $-2017 + 45 = \boxed{-1962}$

x+(2017/x)=x-{x}+(2017/[x]) =>x[x]=2017 Now as x and [x] differ by less than 1 they should be comparable with √2017~44.9 and only such solution is x=-(2017/45). Hence the result is -2017+45=-1972