Problem 264

Solution 1
There are $12 \cdot 11 = 132$ possible situations ($12$ choices for the initially assigned gate, and $11$ choices for which gate Dave's flight was changed to). We are to count the situations in which the two gates are at most $400$ feet apart.
If we number the gates $1$ through $12$, then gates $1$ and $12$ have four other gates within $400$ feet, gates $2$ and $11$ have five, gates $3$ and $10$ have six, gates $4$ and $9$ have have seven, and gates $5$, $6$, $7$, $8$ have eight. Therefore, the number of valid gate assignments is $$2\cdot(4+5+6+7)+4\cdot8 = 2 \cdot 22 + 4 \cdot 8 = 76$$ so the probability is $\frac{76}{132} = \frac{19}{33}$. The answer is $19 + 33 = \boxed{052}$.

Solution 2
As before, derive that there are $132$ possibilities for Dave's original and replacement gates.
Now suppose that Dave has to walk $100k$ feet to get to his new gate. This means that Dave's old and new gates must be $k$ gates apart. (For example, a $100$ foot walk would consist of the two gates being adjacent to each other.) There are $12-k$ ways to pick two gates which are $k$ gates apart, and $2$ possibilities for gate assignments, for a total of $2(12-k)$ possible assignments for each $k$.
As a result, the total number of valid gate arrangements is $$2\cdot 11 + 2\cdot 10 + 2\cdot 9 + 2\cdot 8 = 76$$ and so the requested probability is $\tfrac{19}{33}$ for a final answer of $\boxed{052}$.