I am Floored!

Note that $x > \sqrt[5]{122} \approx 2.614$ and $3 \lfloor{3 \lfloor{3 \lfloor{3 \lfloor{3\rfloor}\rfloor}\rfloor}\rfloor}=3^5 = 243$ . For $x \lfloor{x \lfloor{x \lfloor{x \lfloor{x\rfloor}\rfloor}\rfloor}\rfloor}=122$ $\implies 2.614 < x < 3$ , $\implies \lfloor x \rfloor = 2$ . Then we have:

$\begin{aligned} x \color{#3D99F6} \lfloor{x \lfloor{x \lfloor{x \lfloor{x\rfloor}\rfloor}\rfloor}\rfloor} & = 122 & \small \color{#3D99F6} \text{Let positive integer }n = \lfloor{x \lfloor{x \lfloor{x \lfloor{x\rfloor}\rfloor}\rfloor}\rfloor} \\ (2+\{x\})\color{#3D99F6} n & = 122 \\ 2n + \{x\}n & = 122 \end{aligned}$

Since $2.7 < x < 3$ and $n = \dfrac {122}{2+\{x\}}$ , $\implies 41 \le n \le 45$ . Taking $n=41$ , $\{x\} = \dfrac {122-2n}n = \dfrac {40}{41}$ , $\implies x = \dfrac {122}{41}$ , then we have:

$\begin{aligned} \lfloor{x \lfloor{x \lfloor{x \lfloor{x\rfloor}\rfloor}\rfloor}\rfloor} & = \left \lfloor \frac {122}{41} \left \lfloor \frac {122}{41} \left \lfloor \frac {122}{41} \color{#3D99F6} \left \lfloor \frac {122}{41} \right \rfloor \right \rfloor \right \rfloor \right \rfloor & \small \color{#3D99F6} \text{Note that }\left \lfloor \frac {122}{41} \right \rfloor = 2 \\ & = \left \lfloor \frac {122}{41} \left \lfloor \frac {122}{41} \color{#3D99F6} \left \lfloor \frac {244}{41} \right \rfloor \right \rfloor \right \rfloor & \small \color{#3D99F6} \left \lfloor \frac {244}{41} \right \rfloor = 5 \\ & = \left \lfloor \frac {122}{41} \color{#3D99F6} \left \lfloor \frac {610}{41} \right \rfloor \right \rfloor & \small \color{#3D99F6} \text{and } \left \lfloor \frac {610}{41} \right \rfloor = 14 \\ & = \left \lfloor \frac {1708}{41} \right \rfloor = 41 = n \end{aligned}$

Checking with other values of $n$ , we get $\begin{cases} n=42 & \implies \lfloor{x \lfloor{x \lfloor{x \lfloor{x\rfloor}\rfloor}\rfloor}\rfloor} = 40 \\ n=43 & \implies \lfloor{x \lfloor{x \lfloor{x \lfloor{x\rfloor}\rfloor}\rfloor}\rfloor} =39 \\ n=44 & \implies \lfloor{x \lfloor{x \lfloor{x \lfloor{x\rfloor}\rfloor}\rfloor}\rfloor} =36 \end{cases}$

Therefore, $x = \dfrac {122}{41}$ satisfies the equation and $a+b=122+41 = \boxed {163}$ .

Let's define the functions $f(x)=\lfloor x\lfloor x\lfloor x \lfloor x \rfloor\rfloor\rfloor\rfloor$ and $g(x)=xf(x)$ . Note that $f(x)$ is non-decreasing and $g(x)$ is increasing for $x>0$ . We quickly realize that the $x$ we seek is just below 3, and we compute that $f(2.95)=f(2.99)=41$ . On $I=[2.95,2.99]$ we have $g(x)=41x=122$ for $x=\frac{122}{41}\approx 2.976$ . The answer is $122+41=\boxed{163}$ .

It is clear that $2<x<3$ and after a little playing around it is also pretty obvious that $x$ is pretty close to 3. So let's make a guess that $x=3-\frac{1}{n}=\frac{3n-1}{n}$ .

The solution satisfies $x\lfloor a \rfloor =122$ for some $a$ which implies $\frac{122n}{3n-1}$ is a whole number.

The only values for $n$ are $1$ (which we ignore) and $41$ .

Using this $x=\frac{122}{41}$

Fingers crossed, we try it and it works. As the left side of the equation is increasing for positive x, this is the only solution.