Evaluate $\left\lceil3\left(6-\frac12\right)\right\rceil$.
Firstly, $3\left(6-\frac12\right)=18-1-\frac12=17-\frac12$.  Because $0\le\frac12<1$, we have $\left\lceil17-\frac12\right\rceil=\boxed{17}$.
17
Level 3
Algebra
Problem: Evaluate $\left\lceil3\left(6-\frac12\right)\right\rceil$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To evaluate the expression, first simplify the term inside the parentheses: calculate \(6 - \frac{1}{2}\). This requires converting \(\frac{1}{2}\) into a decimal or a fraction with a common denominator. The value of \(6\) can be expressed as \(\frac{12}{2}\), so \(6 - \frac{1}{2} = \frac{12}{2} - \frac{1}{2} = \frac{11}{2}\).

Next, multiply this result by \(3\): \(3 \left(\frac{11}{2}\right) = \frac{33}{2}\). This fraction can be converted to a decimal for easier evaluation: \(\frac{33}{2} = 16.5\).

Now, apply the ceiling function, which rounds up to the nearest integer. Since \(16.5\) is not an integer, the ceiling function will round it up to \(17\). Therefore, the final answer is \(17\).<|END_OF_LATENT|> The final answer is $17$.