19. (ROM 1) Consider the sequences $\left(a_{n}\right),\left(b_{n}\right)$ defined by  $$ a_{1}=3, \quad b_{1}=100, \quad a_{n+1}=3^{a_{n}}, \quad b_{n+1}=100^{b_{n}} . $$  Find the smallest integer $m$ for which $b_{m}>a_{100}$.
19. Let us define $i_{j}$ for two positive integers $i$ and $j$ in the following way: $i_{1}=i$ and $i_{j+1}=i^{i_{j}}$ for all positive integers $j$. Thus we must find the smallest $m$ such that $100_{m}>3_{100}$. Since $100_{1}=100>27=3_{2}$, we inductively have $100_{j}=10^{100_{j-1}}>3^{100_{j-1}}>3^{3_{j}}=3_{j+1}$ and hence $m \leq 99$. We now prove that $m=99$ by proving $100_{98}<3_{100}$. To do this, we will use the fact that for any $d>12$ (which trivially holds for all $d=100_{i}$), if $c>d^{2}$, then we have 
$$ 
3^{c}>3^{d^{2}}>3^{12 d}=\left(3^{12}\right)^{d}>10000^{d}=\left(100^{d}\right)^{2} 
$$ 
Hence from $3_{3}>\left(100_{1}\right)^{2}$ it inductively follows that $3_{j}>\left(100_{j-2}\right)^{2}>100_{j-2}$ and hence that $100_{99}>3_{100}>100_{98}$. Hence $m=99$.