Find $\left \lceil \frac{\left \lfloor 52/3 \right. \rfloor}{5/23} \right \rceil$.
I know that the symbols $\left \lfloor x \right \rfloor$ and $\left \lceil x \right \rceil$ mean the greatest integer less than or equal to x and the smallest integer greater than or equal to x, respectively. So, to find $\left \lceil \frac{\left \lfloor 52/3 \right. \rfloor}{5/23} \right \rceil$, I need to first evaluate the innermost expression, which is $\left \lfloor 52/3 \right. \rfloor$. This means the greatest integer less than or equal to 52/3, which is 17, since 52/3 is about 17.33 and the next integer is 18, which is too big. So, I can replace $\left \lfloor 52/3 \right. \rfloor$ with 17 and get $\left \lceil \frac{17}{5/23} \right \rceil$. Next, I need to evaluate the fraction 17/(5/23), which is the same as multiplying 17 by 23/5, since dividing by a fraction is the same as multiplying by its reciprocal. So, I get 17 * (23/5) = 391/5, which is 78.2. Now, I need to find the smallest integer greater than or equal to 78.2, which is 79, since 78 is too small. Therefore, the final answer is 79. 