Floor Floor Floor Floor

The approximate value of $x$ will be somewhere around

$\sqrt[4]{41}\approx2.5304$

For $\beta$ ,

Let $x=\frac{m}{n}$ , and $m>n>0$ ,

$\frac{m}{n}k=41$ where $k$ is the integer $\lfloor{x\lfloor{x\lfloor{x}}}\rfloor\rfloor\rfloor$

$k=\frac{41n}{m}$

$m$ must divide $41$ for an integer $k$ , for the smallest positive value, $m=41$

$n$ must be as large as possible, but bear in mind of the approximate value of $x$ . By trial and error, $n=14$ .

Hence, $\beta=\frac{41}{14}$

For $\alpha$ ,

The procedure is similar with the one above, only $n=-17$ .

Hence, $\alpha=-\frac{41}{17}$

$17\alpha+28\beta=41$

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The main idea of the problem is by trial and error . However, next time if you see these kind of problem, it is the best for you to try $n=14$ for $\beta$ , since the problem wants you to get $28\beta$ , clearly is because it needs integer value. For $\alpha$ , it is also the same, substitute $17$ into $n$ immediately.