Evaluate $\cfrac{\left\lceil\cfrac{17}{7}-\left\lceil\cfrac{27}{17}\right\rceil\right\rceil}{\left\lceil\cfrac{27}{7}+\left\lceil\cfrac{7\cdot17}{27}\right\rceil\right\rceil}$
The first thing to be addressed is the fractions under the inner sets of ceiling functions. The smallest integer greater than $\frac{27}{17}$ is $2$. The smallest integer greater than $\frac{7\cdot17}{27}$, which is equal to $\frac{119}{27}$ is $5$. Therefore, the original problem can be rewritten as: \[\frac{\left\lceil\frac{17}{7}-2\right\rceil}{\left\lceil\frac{27}{7}+5\right\rceil}=\frac{\left\lceil\frac{3}{7}\right\rceil}{\left\lceil\frac{62}{7}\right\rceil}\] The smallest integer greater than $\frac{3}{7}$ is $1$ and the smallest integer greater than $\frac{62}{7}$ is $9$. Hence, the final simplified fraction is $\boxed{\frac{1}{9}}$.
\frac{1}{9}
Level 5
Algebra
Problem: Evaluate $\cfrac{\left\lceil\cfrac{17}{7}-\left\lceil\cfrac{27}{17}\right\rceil\right\rceil}{\left\lceil\cfrac{27}{7}+\left\lceil\cfrac{7\cdot17}{27}\right\rceil\right\rceil}$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To evaluate the expression, first calculate the inner fractions. Start with $\cfrac{27}{17}$. This equals approximately 1.588, so $\left\lceil \cfrac{27}{17} \right\rceil = 2$. Next, compute $\cfrac{17}{7}$, which is approximately 2.428. Thus, $\cfrac{17}{7} - 2 = 0.428$, leading to $\left\lceil 0.428 \right\rceil = 1$.

Now, evaluate the denominator. Calculate $\cfrac{27}{7}$, which is approximately 3.857, so $\left\lceil \cfrac{27}{7} \right\rceil = 4$. Next, compute $\cfrac{7 \cdot 17}{27}$. This equals approximately 4.444, and thus $\left\lceil \cfrac{7 \cdot 17}{27} \right\rceil = 5$. Therefore, $\cfrac{27}{7} + 5 = 4 + 5 = 9$, leading to $\left\lceil 9 \right\rceil = 9$.

Now substitute these results back into the original expression: the numerator is 1 and the denominator is 9. Thus, the overall expression simplifies to $\cfrac{1}{9}$. This confirms that the final answer is $\frac{1}{9}$.<|END_OF_LATENT|> The final answer is $\frac{1}{9}$.