Define the [i]hotel elevator cubic [/i]as the unique cubic polynomial $P$ for which $P(11) = 11$, $P(12) = 12$, $P(13) = 14$, $P(14) = 15$. What is $P(15)$?

[i]Proposed by Evan Chen[/i]
1. Define the polynomial \( Q(x) = P(x) - x \). Given that \( P(11) = 11 \) and \( P(12) = 12 \), it follows that \( Q(11) = 0 \) and \( Q(12) = 0 \). Therefore, \( x = 11 \) and \( x = 12 \) are roots of \( Q(x) \).

2. Since \( Q(x) \) is a cubic polynomial and has roots at \( x = 11 \) and \( x = 12 \), we can express \( Q(x) \) as:
   \[
   Q(x) = (x - 11)(x - 12)(ax + b)
   \]
   where \( a \) and \( b \) are constants to be determined.

3. We are also given \( P(13) = 14 \), which implies:
   \[
   Q(13) = P(13) - 13 = 14 - 13 = 1
   \]
   Substituting \( x = 13 \) into the expression for \( Q(x) \), we get:
   \[
   Q(13) = (13 - 11)(13 - 12)(13a + b) = 2 \cdot 1 \cdot (13a + b) = 2(13a + b) = 1
   \]
   Solving for \( 13a + b \):
   \[
   2(13a + b) = 1 \implies 13a + b = \frac{1}{2}
   \]

4. We are also given \( P(14) = 15 \), which implies:
   \[
   Q(14) = P(14) - 14 = 15 - 14 = 1
   \]
   Substituting \( x = 14 \) into the expression for \( Q(x) \), we get:
   \[
   Q(14) = (14 - 11)(14 - 12)(14a + b) = 3 \cdot 2 \cdot (14a + b) = 6(14a + b) = 1
   \]
   Solving for \( 14a + b \):
   \[
   6(14a + b) = 1 \implies 14a + b = \frac{1}{6}
   \]

5. We now have a system of linear equations:
   \[
   \begin{cases}
   13a + b = \frac{1}{2} \\
   14a + b = \frac{1}{6}
   \end{cases}
   \]
   Subtracting the first equation from the second:
   \[
   (14a + b) - (13a + b) = \frac{1}{6} - \frac{1}{2} \implies a = \frac{1}{6} - \frac{3}{6} = -\frac{1}{3}
   \]

6. Substitute \( a = -\frac{1}{3} \) back into the first equation:
   \[
   13\left(-\frac{1}{3}\right) + b = \frac{1}{2} \implies -\frac{13}{3} + b = \frac{1}{2} \implies b = \frac{1}{2} + \frac{13}{3} = \frac{1}{2} + \frac{26}{6} = \frac{1}{2} + \frac{13}{3} = \frac{3}{6} + \frac{26}{6} = \frac{29}{6}
   \]

7. Thus, \( Q(x) = (x - 11)(x - 12)\left(-\frac{1}{3}x + \frac{29}{6}\right) \).

8. To find \( P(15) \), we need \( Q(15) \):
   \[
   Q(15) = (15 - 11)(15 - 12)\left(-\frac{1}{3} \cdot 15 + \frac{29}{6}\right) = 4 \cdot 3 \left(-5 + \frac{29}{6}\right) = 12 \left(-5 + \frac{29}{6}\right)
   \]
   Simplify inside the parentheses:
   \[
   -5 + \frac{29}{6} = -\frac{30}{6} + \frac{29}{6} = -\frac{1}{6}
   \]
   Therefore:
   \[
   Q(15) = 12 \cdot -\frac{1}{6} = -2
   \]

9. Finally, since \( Q(15) = P(15) - 15 \), we have:
   \[
   P(15) - 15 = -2 \implies P(15) = 13
   \]

The final answer is \( \boxed{13} \)