[Algebra]
Problem: Given that $a$ and $b$ are positive and

$$
\lfloor 20-a\rfloor=\lfloor 19-b\rfloor=\lfloor a b\rfloor,
$$

compute the least upper bound of the set of possible values of $a+b$.

Solution: Let the common value of the three expressions in the given equation be $N$. Maximizing $a+b$ involves making at least one of $a$ and $b$ somewhat large, which makes the first two expressions for $N$ small. So, to maximize $a+b$, look for the least possible value of $N$. One can show that $N=14$ is not possible because that would require $a>5$ and $b>4$, which implies $a b>20$. But $N=15$ is possible by setting $a=4+x, b=3+y$, where $0<x, y \leq 1$. The goal is to find the least upper bound for $x+y$ given $15 \leq(4+x)(3+y)<16 \Rightarrow 3 \leq 3(x+y)+y+x y<4$. This is equivalent to seeking the maximum value of $x+y$ given $3(x+y)+y+x y \leq 4$. By inspection, if $x=1$ and $y=\frac{1}{5}$, then $3(x+y)+y+x y=4 \leq 4$. This is in fact optimal. To see this, consider that because $3 x+3 y+y+x y \leq 4$, it follows that $y \leq \frac{4-3 x}{x+4}$, and so $x+y \leq x+\frac{4-3 x}{x+4} \leq \frac{x^{2}+x+4}{x+4}$, which is increasing on $0 \leq x \leq 1$. Thus the maximum for $x+y$ is attained when $x=1$. Hence the least upper bound for $a+b$ is $5+\left(3+\frac{1}{5}\right)=\frac{\mathbf{4 1}}{\mathbf{5}}$.

Final answer: $\frac{41}{5}$