Equating Floors

For $x > 0$ , $\left \lfloor \dfrac x8 \right \rfloor = \left \lfloor \dfrac x{11} \right \rfloor = k$ , where $k$ is a non-negative integer. Then we have:

$\begin{cases} \left \lfloor \dfrac x8 \right \rfloor = k & \implies 8k \le x < 8(k+1) \implies 8k \le x \le 8k + 7 \\ \left \lfloor \dfrac x{11} \right \rfloor = k & \implies 11k \le x < 11(k+1) \implies 11k \le x \le 11k + 10 \end{cases}$

Since $8k \le 11k$ and $8k + 7 \le 11k + 10$ , therefore, $11k \le x \le 8k+7$ .

\(\begin{array} {} k = 0 & \implies 11(0) \le x \le 8(0)+7 & \implies x = 1,2,3,4,5,6,7 & \color{blue}{\text{7 solutions}} \\ k = 1 & \implies 11(1) \le x \le 8(1)+7 & \implies x = 11,12,13,14,15 & \color{blue}{\text{5 solutions}} \\ k = 2 & \implies 11(2) \le x \le 8(2)+7 & \implies x = 22,23 & \color{blue}{\text{2 solutions}} \\ k = 3 & \implies 11(3) \le x \le 8(3)+7 & \implies \require{cancel} \color{red}{\cancel{33 \le x \le 31}} & \color{red}{\text{No solution for }k \ge 3} \end{array} \)

Therefore, the total number of integer solutions is $7+5+2=\boxed{14}$ .

Here,

$\left\lfloor \dfrac{x}{8} \right\rfloor = \left\lfloor \dfrac{x}{11} \right\rfloor$

Since, $x\in\mathbb{Z^{+}}$ , so we can have,

$\left\lfloor \dfrac{x}{8} \right\rfloor = \left\lfloor \dfrac{x}{11} \right\rfloor = 0 \\ \left\lfloor \dfrac{x}{8} \right\rfloor = \left\lfloor \dfrac{x}{11} \right\rfloor =1 \\ \left\lfloor \dfrac{x}{8} \right\rfloor = \left\lfloor \dfrac{x}{11} \right\rfloor = 2$

These are the only three positive integral values for $x$ because the smallest value of $x$ for which $\left\lfloor \dfrac{x}{8} \right\rfloor = 3$ is $24$ and smallest value of $x$ for which $\left\lfloor \dfrac{x}{11} \right\rfloor = 3$ is $33$ , but at $x=33$ , $\left\lfloor \dfrac{x}{8} \right\rfloor = 4$ , which means we cannot have a common $x$ for $\left\lfloor \dfrac{x}{8} \right\rfloor = \left\lfloor \dfrac{x}{11} \right\rfloor = 3$ and this trend continues

$\forall$ $x\ge 3$ where $x\in\mathbb{Z^{+}}$ .

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So, we have three cases:

$\underline{\text{Case 1:}}~~~ \left\lfloor \dfrac{x}{8} \right\rfloor = \left\lfloor \dfrac{x}{11} \right\rfloor = 0$

In this case $0<x<8$ .

So $x\in\left[1,2,3,4,5,6,7\right]$ .

$\blacksquare$ Total values of $x = 7$ .

$\underline{\text{Case 2:}}~~~ \left\lfloor \dfrac{x}{8} \right\rfloor = \left\lfloor \dfrac{x}{11} \right\rfloor = 1$

In this case $11\le x < 16$ , because for $8\le x < 11$ , $\left\lfloor \dfrac{x}{8} \right\rfloor = 1$ , while $\left\lfloor \dfrac{x}{11} \right\rfloor = 0$ . Also, at $16\le x < 22$ , $\left\lfloor \dfrac{x}{8} \right\rfloor = 2$ , while $\left\lfloor \dfrac{x}{11} \right\rfloor = 1$ .

So $x\in\left[11,12,13,14,15\right]$ .

$\blacksquare$ Total values of $x = 5$ .

$\underline{\text{Case 3:}}~~~ \left\lfloor \dfrac{x}{8} \right\rfloor = \left\lfloor \dfrac{x}{11} \right\rfloor = 2$

In this case $22 \le x \le 23$ or simply $x\in\left[22,23\right]$ , because for $x \ge 24$ , $\left\lfloor \dfrac{x}{8} \right\rfloor \neq \left\lfloor \dfrac{x}{11} \right\rfloor$ (as deduced earlier) and for $16\le x < 22$ , $\left\lfloor \dfrac{x}{8} \right\rfloor = 2$ , while $\left\lfloor \dfrac{x}{11} \right\rfloor = 1$ as already shown in previous case.

So $x\in\left[22,23\right]$ .

$\blacksquare$ Total values of $x = 2$ .

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Hence, the required answer: $7+5+2=\boxed{14}$

Let $\left \lfloor \dfrac x{n_1 }\right \rfloor = \left \lfloor \dfrac x{n_2} \right \rfloor = k,~n_2>n_1$

So, solution for $x$ becomes $x\ge n_2k~~\text{and}~ ~x\le n_1(k+1)-1$

So, for the set to be not null, $n_2k\le n_1(k+1)-1\\k\le \dfrac {n_1-1}{n_2-n_1}$

Substituting values, we get $k<3$

So, for $k=0\to \color{#3D99F6}{7~\text{values}}\\k=1\to \color{#3D99F6}{5~\text{values}}\\k=2\to \color{#3D99F6}{2~\text{values}}$

Hence $\color{#3D99F6}{\boxed{14~\text{values}}}$