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.

Chimes per full day. Each of the $24$ half-hours gives $1$ chime, and the hour chimes run $1, 2, \dots, 12$ twice, for $2 \cdot 78 = 156$. A full day has $24 + 156 = 180$ chimes.

Rest of February 26. After $11{:}15$ AM the half-hour chimes occur at $11{:}30$ AM, $12{:}30$ PM, ..., $11{:}30$ PM: $13$ of them. The hour chimes are noon ($12$) and $1$ PM through $11$ PM ($1 + \cdots + 11 = 66$): $78$ in all. Total for the partial day: $13 + 78 = 91$.

Counting forward. Still needed after February 26: $2003 - 91 = 1912$. Since $1912 = 10 \cdot 180 + 112$, ten full days (February 27, February 28, March 1 through March 8) bring the count to $91 + 1800 = 1891$, and the remaining $112$ chimes occur during the eleventh day. That day is March 9.

(The midnight chimes belong to the start of the new day, but even if assigned to the previous day the remainder is still positive and the eleventh day is unchanged.)

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