problem: Find $\left \lceil \frac{\left \lfloor 52/3 \right. \rfloor}{5/23} \right \rceil$.
solution: Since $17 = \frac{51}{3} < \frac {52}3 < \frac {54}3 = 18$, the floor  of $52/3$ is $17$. The given quantity is thus equal to $$\left \lceil \frac{17}{5/23} \right \rceil = \left \lceil \frac{391}{5} \right \rceil = \left \lceil 78.2 \right. \rceil = \boxed{79}.$$