2003 AMC 10B Problem 22

Step 1: chimes in one full day.

1. Half-hour chimes: one every half hour past the hour, $24$ per day.
2. Hour chimes: the clock strikes $12$ at midnight, then $1, 2, \ldots, 11$ in the morning, then $12$ at noon and $1, 2, \ldots, 11$ in the afternoon and evening. So the numbers $1$ to $12$ each come up twice. Since $1 + 2 + \cdots + 12 = 78$, that is $2 \times 78 = 156$ chimes.
3. One full day: $24 + 156 = 180$ chimes.

Step 2: chimes on the rest of February 26 (after 11:15 AM).

1. Half-hour chimes: 11:30 AM, then 12:30 PM, 1:30 PM, ..., 11:30 PM. That is $1 + 12 = 13$ chimes.
2. Hour chimes: noon gives $12$, then 1 PM through 11 PM give $1 + 2 + \cdots + 11 = 66$. That is $78$.
3. So February 26 has $13 + 78 = 91$ chimes. (Midnight belongs to the start of February 27.)

Step 3: count forward.

1. Chimes still needed after February 26: $2003 - 91 = 1912$.
2. Divide by a full day: $1912 = 10 \times 180 + 112$. So ten full days pass, and $112$ more chimes are needed on the next day.
3. The ten full days are February 27, February 28 (2003 is not a leap year), and March 1 through March 8. The $2003^{\text{rd}}$ chime comes during the eleventh day, March 9. Since $112$ is less than $180$, it does fall within that day.

The answer is $\boxed{\textbf{(B)}\ \text{March 9}}$.

For something that repeats every day, find the amount per day, handle the uneven first day on its own, and then divide with remainder. A leftover that is more than zero means the target falls inside the next day, so you count one more day than the quotient. The last trap is the calendar: February 2003 has $28$ days.