2018 AMC 10A Problem 14

Let $a=3^{96}$ and $b=2^{96}$. The expression is $\dfrac{81a+16b}{a+b}=16+\dfrac{65a}{a+b}$, so it is less than $16+65=81$. To show the floor is $80$, we also need the expression to be greater than $80$. This is equivalent to $81a+16b>80a+80b$, or $a>64b$. Because $\left(\dfrac32\right)^2=\dfrac94>2,$ we have $\dfrac{a}{b}=\left(\dfrac32\right)^{96}>2^{48}>64.$ Hence the expression is greater than $80$ and less than $81.$ Thus, A is the correct answer.