2016 AMC 10A Problem 13

Several people move, but they always fill the same five seats. That gives a total that cannot change.

1. Before and after, the five friends sit in seats $1, 2, 3, 4, 5$. So the seat numbers add up to $15$ both times.
2. Since the total stays $15$, all the changes in seat number must add up to $0$.
3. List the changes (moving right adds, moving left subtracts):
- Bea: $+2$.
- Ceci: $-1$.
- Dee and Edie swapped. If one gained some seats, the other lost exactly as many, so together they add $0$.
- Ada: unknown.
4. So $2 - 1 + 0 + (\text{Ada's change}) = 0$, and Ada's change is $-1$. She ended up one seat to the left of where she started.
5. Ada ended in an end seat, $1$ or $5$. If she ended in seat $5$ she would have started in seat $6$, which does not exist. So she ended in seat $1$ and started in seat $2$.
6. Check that this can really happen. Before: Bea $1$, Ada $2$, Ceci $3$, Dee $4$, Edie $5$. After: seat $1$ empty for Ada, Ceci $2$, Bea $3$, Edie $4$, Dee $5$. Every clue holds.

The answer is $\boxed{\textbf{(B)}\ 2}$.

When people shuffle among a fixed set of seats, the sum of the seat numbers never changes, so all the moves must cancel out. That single fact finds the one unknown move without trying cases. Look for a total that stays the same whenever a puzzle describes many moves and asks about one of them.

Look only at the other four friends. Their moves add up to $2 - 1 + 0 = +1$.

• If Ada ends in seat $5$, the other four end in seats $1$ to $4$, which add to $10$. Before, they sat in every seat except Ada's, adding to $15 - (\text{Ada's seat})$. Their moves raised the total by $1$, so $15 - (\text{Ada's seat}) + 1 = 10$, and Ada's seat would be $6$. Impossible.
• If Ada ends in seat $1$, the other four end in seats $2$ to $5$, which add to $14$. Now $15 - (\text{Ada's seat}) + 1 = 14$, so Ada started in seat $2$.