A school is arranging chairs in rows for an assembly. $11$ chairs make a complete row, and right now there are $110$ chairs total. The school wants to have as few empty seats as possible, but all rows of chairs must be complete. If $70$ students will attend the assembly, how many chairs should be removed?
The original number of chairs is divisible by $11$, and the final number of remaining chairs must also be divisible by $11$ to have complete rows. This means that the number of chairs removed must be divisible by $11$ because the difference between two multiples of $b$ is also a multiple of $b$. In other words, $$b \cdot m - b \cdot n = b \cdot k.$$ The smallest multiple of $11$ greater than $70$ is $77$, so we want $77$ chairs to remain. This means we must take away \begin{align*}
110 - 77 &= 11 \cdot 10 - 11 \cdot 7 \\
&= 11(10-7) \\
&= 11 \cdot 3 \\
&= \boxed{33}
\end{align*} chairs. We check that $33$ is divisible by $11$, and it is, so we took away $3$ complete rows of chairs.
33
Level 3
Prealgebra
Problem: A school is arranging chairs in rows for an assembly. $11$ chairs make a complete row, and right now there are $110$ chairs total. The school wants to have as few empty seats as possible, but all rows of chairs must be complete. If $70$ students will attend the assembly, how many chairs should be removed?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine how many chairs should be removed, first calculate the number of complete rows that can be formed with the available chairs. With 110 chairs and 11 chairs per row, the total number of rows is 110 divided by 11, which equals 10 rows. 

Next, consider the number of students attending the assembly, which is 70. To minimize empty seats while ensuring all rows are complete, we need to find the number of rows that can accommodate 70 students. Dividing 70 by 11 gives approximately 6.36, meaning we can have a maximum of 6 complete rows. 

Calculating the number of chairs needed for 6 rows, we multiply 6 rows by 11 chairs per row, resulting in 66 chairs. 

Now, to find out how many chairs need to be removed, subtract the number of chairs needed (66) from the total number of chairs (110). This gives us 110 minus 66, which equals 44 chairs. 

However, we need to ensure that the number of empty seats is minimized. If we consider using 7 rows instead, we would have 7 rows times 11 chairs per row, which equals 77 chairs. This would leave us with 70 students and 7 empty seats, which is better than the 44 chairs removed scenario. 

Thus, to achieve the optimal arrangement with the least number of empty seats, we should remove 33 chairs from the original 110, leaving us with 77 chairs for 70 students. This results in 7 empty seats, which is the most efficient configuration. 

Therefore, the final answer is 33.<|END_OF_LATENT|> The final answer is $33$.