Listen to the note!

Using computation by rounding , we have the $n$ th Fibonacci number $F_n = \left \lceil \dfrac {\varphi^n}{\sqrt 5} \right \rfloor$ , where $\varphi = \dfrac {1+\sqrt 5}2$ is the golden ratio. Then $A = F_{49} = \left \lceil \dfrac {\varphi^{49}}{\sqrt 5} \right \rfloor = \left \lceil \dfrac {(1.618...)^{49}}{\sqrt 5} \right \rfloor < 1.7^{49} = B$ , $\implies \boxed{B > A}$ .

I refused to use the note!

It's easy to show by hand that $17^4 > 10^{5}$ . Equivalently, $\begin{aligned} 17^{48} &>& 10^{60} \\ 17^{49} > 17^{48} &>& 10^{60} > 10^{59} \\ 9\times 17^{49} &>& 9\times 10^{59} \\ 9\times 17^{49} &>& 9\times 10^{59} > 8\times 10^{59} \\ 9\times 17^{49} &>& 8\times 10^{59} \\ B=\left( \frac{17}{10}\right)^{49} &>& \frac89 \times 10^{11} \\ B &>& \frac89 \times 10^{11} > A \end{aligned}$