Evaluate $\lfloor0.999\rfloor+\lceil2.001\rceil$.
The greatest integer that is less than or equal to $0.999$ is $0$, so $\lfloor0.999\rfloor=0$. The smallest integer that is greater than or equal to $2.001$ is $3$, so $\lceil2.001\rceil=3$. Therefore, $\lfloor0.999\rfloor+\lceil2.001\rceil=0+3=\boxed{3}$.
3
Level 1
Algebra
Problem: Evaluate $\lfloor0.999\rfloor+\lceil2.001\rceil$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To evaluate the expression, first consider the floor function applied to 0.999. The floor function, denoted as $\lfloor x \rfloor$, returns the greatest integer less than or equal to $x$. Since 0.999 is less than 1 but greater than 0, we have $\lfloor 0.999 \rfloor = 0$.

Next, consider the ceiling function applied to 2.001. The ceiling function, denoted as $\lceil x \rceil$, returns the smallest integer greater than or equal to $x$. Since 2.001 is slightly greater than 2, we find that $\lceil 2.001 \rceil = 3$.

Now, we combine the results of the two functions: $\lfloor 0.999 \rfloor + \lceil 2.001 \rceil = 0 + 3$. 

Performing the addition yields $3$. 

Thus, the final answer is $3$.<|END_OF_LATENT|> The final answer is $3$.