# Difference between revisions of "2003 AMC 10B Problems/Problem 22"

## Problem

A clock chimes once at $30$ minutes past each hour and chimes on the hour according to the hour. For example, at $1 \text{PM}$ there is one chime and at noon and midnight there are twelve chimes. Starting at $11:15 \text{AM}$ on $\text{February 26, 2003},$ on what date will the $2003^{\text{rd}}$ chime occur? $\textbf{(A) } \text{March 8} \qquad\textbf{(B) } \text{March 9} \qquad\textbf{(C) } \text{March 10} \qquad\textbf{(D) } \text{March 20} \qquad\textbf{(E) } \text{March 21}$

## Solution

First, find how many chimes will have already happened before midnight (the beginning of the day) of $\text{February 27, 2003}.$ $13$ half-hours have passed, and the number of chimes according to the hour is $1+2+3+\cdots+12.$ The total number of chimes is $13+78=91$

Every day, there will be $24$ half-hours and $2(1+2+3+\cdots+12)$ chimes according to the arrow, resulting in $24+156=180$ total chimes.

On $\text{February 27},$ the number of chimes that still need to occur is $2003-91=1912.$ $1912/180=10\text{r}112.$ Round up, and it is $11$ days past $\text{February 27},$ which is $\boxed{\textbf{(B) \ } \text{March 9}}$

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