2003 AMC 10B Problem 22

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

Periodic processes are handled by finding the count per period, peeling off the irregular first stretch, and using division with remainder. A positive remainder means the target lands strictly inside the next period, so you round the quotient up. The calendar detail (28 days in February 2003) is the last place to slip.

• Sliding the date by one: forgetting February 2003 has $28$ days, or counting the tenth full day as the answer instead of the day the remainder falls on.
• Using a $24$-hour count for the hour chimes (e.g. $1 + \cdots + 24$) instead of the $12$-hour cycle repeated twice.