What is the greatest integer $n$ such that $n^2 - 11n +24 \leq 0$?
We can factor $n^2-11n+24$ as $(n-3)(n-8)$. For this quantity to be less than or equal to 0, one of the factors must be less than or equal to 0 and the other factor must be greater than or equal to 0. Specifically, since $n-8<n-3$ for all $n$, we must have $$n-8 \le 0 \le n-3.$$ The first inequality, $n-8\le 0$, tells us that $n\le 8$. The second inequality, $0\le n-3$, tells us that $n\ge 3$. The solutions to the original inequality must satisfy both conditions, so they are given by $3\le n\le 8$. The largest integer in this interval is $n=\boxed{8}$.
8
Level 3
Algebra
Problem: What is the greatest integer $n$ such that $n^2 - 11n +24 \leq 0$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the inequality \( n^2 - 11n + 24 \leq 0 \), we first need to find the roots of the corresponding quadratic equation \( n^2 - 11n + 24 = 0 \). We can apply the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -11 \), and \( c = 24 \).

Calculating the discriminant:
\[
b^2 - 4ac = (-11)^2 - 4 \cdot 1 \cdot 24 = 121 - 96 = 25.
\]
Since the discriminant is positive, there are two distinct real roots. Now we compute the roots:
\[
n = \frac{11 \pm \sqrt{25}}{2} = \frac{11 \pm 5}{2}.
\]
This gives us:
\[
n_1 = \frac{16}{2} = 8 \quad \text{and} \quad n_2 = \frac{6}{2} = 3.
\]

Next, we analyze the sign of the quadratic expression \( n^2 - 11n + 24 \) in the intervals determined by the roots \( n = 3 \) and \( n = 8 \). The critical points divide the number line into three intervals: \( (-\infty, 3) \), \( (3, 8) \), and \( (8, \infty) \).

1. For \( n < 3 \), choose \( n = 0 \):
   \[
   0^2 - 11 \cdot 0 + 24 = 24 > 0.
   \]
2. For \( 3 < n < 8 \), choose \( n = 5 \):
   \[
   5^2 - 11 \cdot 5 + 24 = 25 - 55 + 24 = -6 < 0.
   \]
3. For \( n > 8 \), choose \( n = 9 \):
   \[
   9^2 - 11 \cdot 9 + 24 = 81 - 99 + 24 = 6 > 0.
   \]

The quadratic expression is negative or zero in the interval \( [3, 8] \). Therefore, the greatest integer \( n \) satisfying the inequality \( n^2 - 11n + 24 \leq 0 \) is the upper endpoint of this interval, which is \( 8 \).<|END_OF_LATENT|> The final answer is $8$.