Fantastic integers

Let us define a transitional fraction $\{x_+\}$ as the smallest fraction where the value of $\lfloor x \rfloor \{x_+\}$ increases by $1$ . Transitional fraction is useful as it is only at this fractional value that the product of integral part and fractional part changes value and we can use it in solving this problem.

Starting from the inner-most floor function. Since $3^4 < 88 < 4^4$ , we know that $3 < x < 4$ .

$f(x) = \lfloor x \lfloor x \lfloor x \lfloor x \rfloor \rfloor \rfloor \rfloor = \lfloor x \lfloor x \lfloor 3x \rfloor \rfloor \rfloor = 88$

Case 1: $\lfloor x \lfloor x \lfloor 3(3+\{ x_+ \}) \rfloor \rfloor \rfloor \quad \Rightarrow \{x_+\} = \frac{1}{3} \\ \Rightarrow f\left(\frac{10}{3}\right) = \lfloor \frac{10}{3} \lfloor \frac{10}{3} \lfloor 10 \rfloor \rfloor \rfloor = \lfloor \frac{10}{3} \lfloor \frac{100}{3} \rfloor \rfloor = \lfloor \frac{330}{3} \rfloor = 110 > 88 \\ \Rightarrow x < \frac{10}{3}$

Case 2: $\lfloor x \lfloor 9(3+\{ x_+ \}) \rfloor \rfloor \Rightarrow \{x_+\} = \frac{1}{9} \\ \Rightarrow f\left(\frac{28}{9}\right) = \lfloor \frac{28}{9} \lfloor 28 \rfloor \rfloor = \lfloor \frac{784}{9} \rfloor = 87 < 88 \quad \Rightarrow x > \frac{28}{9} \\ \text{If } \{x_+\} = \frac{2}{9} \quad \Rightarrow f\left(\frac{29}{9}\right) = 93 > 88 \quad \Rightarrow \frac{28}{9} < x < \frac{29}{9} \\ \text{Assuming } \{x_+\} = \frac{1}{6} \quad \Rightarrow f\left(\frac{19}{6}\right) = 88 \quad \Rightarrow x \approx \frac{19}{6}$

$\Rightarrow \lfloor 7x \rfloor = \lfloor \frac{7\times 19}{6} \rfloor = \lfloor \frac{133}{6} \rfloor = \boxed{22}$

$\lfloor x \lfloor x \lfloor x \lfloor x \rfloor \rfloor \rfloor \rfloor = 88$

We can observe that $x^4$ would be near $88$ and the closest value of $x$ for this is $3$ .

To minimize our search even more, we can try taking $x$ as $3.05, 3.10, 3.15, 3.20$

We would observe that $3.15, 3.16, 3.17$ satisfy the condition.

Any of the three as $x$ will give $\lfloor{7x}\rfloor = \boxed {22}$