2007 AMC 10B Problem 9

Two jobs: find which appearance of $s$ the last one is, then work out how far it moves.

1. Count the $s$'s word by word:

The last $s$ is the $12$th one.
2. The $12$th appearance moves $1 + 2 + 3 + \cdots + 12$ places. Pair the numbers from the outside in: $1 + 12$, $2 + 11$, and so on, six pairs of $13$. So the shift is
$$6 \times 13 = 78 \text{ places}.$$
3. The alphabet has $26$ letters, and moving $26$ places brings a letter back to itself. Since $78 = 3 \times 26$, the letter goes around three full times and stops where it started.
4. So the last $s$ becomes $s$.

The answer is $\boxed{\textbf{(D)}\ s}$.

With a wrap-around alphabet, only the leftover after taking out full trips of $26$ matters. Figure out the total shift first and remove the full trips, rather than stepping letter by letter. Most of the danger in this problem is simply counting the $s$'s correctly, so a table is worth the few seconds.

First count how many times the letter $s$ appears, word by word:

Running total: $1, 3, 4, 4, 8, 10, 12$. The final $s$ is the $12$th occurrence.

The $k$th occurrence of a letter moves $1 + 2 + \cdots + k = \frac{k(k+1)}{2}$ places to the right. For $k = 12$ this is $\frac{12 \cdot 13}{2} = 78$.

Since the alphabet has $26$ letters, a shift of $78 = 3 \cdot 26$ brings the letter all the way around three full times and back to where it started. The last $s$ is replaced by $s$ itself.

The answer is $\boxed{\textbf{(D)}\ s}$.