So I was yet again looking on Michael Penn's Youtube channel to see if there were any math equations on his channel that I thought that I might be able to solve when I came across this member only video of his that I found interesting (don't worry, I also don't have access to it). The equation on the thumbnail was$$\lfloor2x\rfloor+\lfloor3x\rfloor+\lfloor7x\rfloor=2008$$which I thought that I might be able to solve. Here is my attempt at doing so:$$\begin{align}f(x)=\lfloor2x\rfloor&+\lfloor3x\rfloor+\lfloor7x\rfloor=2008\\&\approx\lfloor x\rfloor(2+3+7)\\&=12\lfloor x\rfloor=2008\end{align}$$Now, dividing by $12$ on both sides gets us$$\lfloor x\rfloor=167+\dfrac23\implies\lfloor\lfloor x\rfloor\rfloor=\lfloor x\rfloor=167\implies x\in[167,168)$$If you're confused on the last bit, it basically means that $x$ must be a number in a set (let's say set $\mathbb A$ for example) that are all numbers in $\mathbb R$ that are greater than $167$ but are less than $168$, with $167$ included in said set.
Now, let $x=a_{n,k}=167+f(n,k)$ where $f(n,k)=\dfrac{n'}{k'},2\leq n\lt k,3\leq k\require{cancel}\cancel\implies\dfrac{\dfrac{df}{dn}}{\dfrac{df}{dk}}$ however it means that any term $n_\alpha$ and $k_\beta$ where $\alpha,\beta$ are in the set of integers (with $\alpha<\beta$). Now let's have there being a set containing all of the prime numbers. (Let's call this set $\mathbb B$) Starting with $\alpha=1,\beta=2$ (so $n_1=2,k_2=3$) we let $\alpha\longrightarrow \beta-1$, and every time we hit $\alpha=\beta-1$, we let $\beta\mapsto\beta+1$ until $167+\dfrac{n_\alpha}{k_\beta}$ is a solution to $f(x)=2008$, so we have$$a_{1,2}\implies x=167+\dfrac23\implies f(x)=2011$$$$a_{1,3}\implies x=167.4\implies f(x)=2007$$$$a_{2,3}\implies x=167.6\implies f(x)=2010$$$$a_{2,4}\implies x=167+\dfrac37\implies f(x)=2008$$Which is actually our lower limit of the solution, as confirmed by Wolfram Alpha. We can actually find our upper limit by allowing $n$ and $k$ to be non-prime numbers (however in the set of natural numbers this time), while still having $n\leq k-1$, starting with $n=1$, $k=2$:$$f(167.5)=200\color{red}{9}\text{ however }\lim_{x\longrightarrow167.5^-}f(x)=2008$$which means the solution is represented by the inequality $\dfrac{1172}7\leq x\lt\dfrac{335}2$, which implies that our solution set is represented by$$x\in\left[\dfrac{1772}7,167.5\right)$$
My question
Is there anyway I could solve this more quickly, or is this already the most efficient solution possible?
I absolutely don't understand your notations (and $\approx$ doesn't mean anything), but your solution of $\frac{1172}{7} \leqslant x < \frac{335}{2}$ seems to be the right answer. If you want an other method, here is one. Let us call $(1)$ this equation and $A$ the set of all real solutions. Change your unknown with $y = 42x = 2 \cdot 3 \cdot 7 \cdot x$. The equation becomes, $\lfloor y/21 \rfloor + \lfloor y/14 \rfloor + \lfloor y/6 \rfloor = 2008$, let us call this equation $(2)$.
Notice that if $n$ is an integer and $y$ a real number such that $n \leqslant y < n + 1$, and $m > 0$ an integer, we have, $\frac{n}{m} \leqslant \frac{y}{m} < \frac{n + 1}{m}$. It implies that $\left\lfloor \frac{n}{m} \right\rfloor \leqslant \left\lfloor \frac{y}{m} \right\rfloor$.
Reciprocally, assume that $p$ is an integer such that $\frac{n}{m} < p \leqslant \frac{y}{m}$. Then $p < \frac{n + 1}{m}$ so $n < pm < n + 1$, which is impossible because $pm$ is an integer. It proves the equality $\left\lfloor \frac{n}{m} \right\rfloor = \left\lfloor \frac{y}{m} \right\rfloor$.
As $n = \lfloor y \rfloor$, it can be rewritten as $\left\lfloor \frac{\lfloor y \rfloor}{m} \right\rfloor = \left\lfloor \frac{y}{m} \right\rfloor$, for any real number $y$ and any integer number $m > 0$. Using this, we have for all $y$ that $y$ solution of the equation $(2)$ if and only if its floor is.
As $y \mapsto \lfloor y/21 \rfloor + \lfloor y/14 \rfloor + \lfloor y/6 \rfloor$ is an increasing function that tends to $\pm\infty$ when $y \rightarrow \pm\infty$, thus the set $S = \{n \in \mathbb{Z}|n$ is solution of $(2)\}$ is an interval of integers (or empty). Let $S = [\![a,b]\!]$ for $a \leqslant b$ (if $S \neq \emptyset$). The previous statement implies that the set of all real solution of $(2)$ is the interval $I = [a,b + 1[$ (or $\emptyset$ if $S = \emptyset$).
Now, notice that, $$ \lfloor 7031/21 \rfloor + \lfloor 7031/14 \rfloor + \lfloor 7031/6 \rfloor = 2007, \quad \lfloor 7032/21 \rfloor + \lfloor 7032/14 \rfloor + \lfloor 7032/6 \rfloor = 2008, $$ $$ \lfloor 7034/21 \rfloor + \lfloor 7034/14 \rfloor + \lfloor 7034/6 \rfloor = 2008, \quad \lfloor 7035/21 \rfloor + \lfloor 7035/14 \rfloor + \lfloor 7035/6 \rfloor = 2009. $$ Therefore, $S \neq \emptyset$ and $a = 7032$ and $b = 7034$. It implies that $I = [7032,7035[$ so $A = \left[\frac{7032}{42},\frac{7035}{42}\right[ = \left[\frac{1172}{7},\frac{735}{2}\right[$.