AMC 10 · 2003 · #22
Easy mode Grade 4A clock chimes one time at 30 minutes past every hour. On the hour, it chimes the hour number: one chime at 1 o'clock, and twelve chimes at 12 o'clock (this happens at both noon and midnight). You start counting at 11:15 AM on February 26, 2003. On what date will the 2003rd chime happen?
Pick an answer.
AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A clock chimes once at every half hour, and on each hour it chimes a number of times equal to the hour on a 12-hour dial (so 1 through 12). Starting the count at 11:15 AM on February 26, 2003, find the calendar date on which the 2003rd chime happens.
Givens: At $30$ minutes past every hour the clock chimes exactly once.; On the hour it chimes the hour number: $1$ at 1:00, up to $12$ at 12:00 (noon and midnight both give $12$).; The counting starts at 11:15 AM on February 26, 2003.; 2003 is not a leap year, so February has $28$ days.; Answer choices: (A) March 8, (B) March 9, (C) March 10, (D) March 20, (E) March 21.
Unknowns: The date on which the 2003rd chime occurs.
Understand
Restated: A clock chimes once at every half hour, and on each hour it chimes a number of times equal to the hour on a 12-hour dial (so 1 through 12). Starting the count at 11:15 AM on February 26, 2003, find the calendar date on which the 2003rd chime happens.
Givens: At $30$ minutes past every hour the clock chimes exactly once.; On the hour it chimes the hour number: $1$ at 1:00, up to $12$ at 12:00 (noon and midnight both give $12$).; The counting starts at 11:15 AM on February 26, 2003.; 2003 is not a leap year, so February has $28$ days.; Answer choices: (A) March 8, (B) March 9, (C) March 10, (D) March 20, (E) March 21.
Plan
Primary tool: #7 Identify Subproblems
Secondary: #2 Make a Systematic List, #8 Analyze the Units
Counting to $2003$ chime by chime is hopeless, so Tool #7 (Identify Subproblems) splits the job into four clean pieces: (1) how many chimes a normal full day holds, (2) how many chimes are left in the first partial day after 11:15 AM, (3) how many chimes remain to reach $2003$ and how many whole days that buys, and (4) which calendar date the leftover chimes land on. Tool #8 (Analyze the Units) turns the clock into a steady rate of chimes-per-day so the big number can be divided rather than counted, and Tool #2 (Make a Systematic List) keeps the hour-by-hour tally on the partial day honest.
Execute — Answer: B
4.NBT.B.5 Step 1 Chimes in one full day
- A full day has two kinds of chimes.
- The half-hour chimes: there are $24$ half-hours in a day, each giving one chime, for $24$ chimes.
- The on-the-hour chimes: the 12-hour dial runs $1,2,\dots,12$, which sums to $1+2+\cdots+12 = 78$, and this happens twice a day (AM and PM), for $2 \times 78 = 156$ chimes.
- So a full day holds $24 + 156 = 180$ chimes.
💡 Split the day into its steady half-hour ticks and its climbing hourly counts, then add the two totals.
4.OA.A.3 Step 2 Chimes left on the first day
- After 11:15 AM on Feb 26, list every chime through the end of that day.
- Half-hour chimes at 11:30 AM, 12:30, 1:30, ..., 11:30 PM are $13$ single chimes.
- On-the-hour chimes from noon through 11 PM are $12$ (noon) plus $1+2+\cdots+11 = 66$, for $78$.
- That is $13 + 78 = 91$ chimes on the remainder of Feb 26.
💡 The starting time falls mid-day, so hand-count just the chimes left before the next midnight.
4.OA.A.3 Step 3 How many chimes remain
- The first $91$ chimes are used up on Feb 26.
- The chimes still needed to reach $2003$ are $2003 - 91 = 1912$, and every one of these happens on Feb 27 or later, where each full day contributes exactly $180$.
💡 Peel off the messy partial day first, so what is left is a whole number of clean full days plus a bit.
4.NBT.B.6 Step 4 Divide into whole days plus a remainder
- Divide the remaining chimes by the daily total: $1912 \div 180 = 10$ with remainder $112$.
- So $10$ complete days use up $1800$ chimes, and $112$ chimes are still needed after that.
- Because the remainder is not zero, the 2003rd chime does not fall on the last full day — it falls partway through the very next day.
💡 Ten whole days fit, and the leftover 112 chimes spill into one more day — that extra day is where the target lands.
4.OA.A.3 Step 5 Count the days forward
- The $10$ full days start on Feb 27.
- Counting: Feb 27, Feb 28 (2 days), then March 1, 2, 3, 4, 5, 6, 7, 8 (8 more) — that is exactly $10$ full days, ending March 8.
- The remaining $112$ chimes happen on the next day, which is March 9.
- So the 2003rd chime occurs on March 9, choice (B).
💡 Ten days after Feb 27 ends on March 8, so the spillover chimes belong to March 9.
4.NBT.B.5 A full day has two kinds of chimes. The half-hour chimes: there are $24$ half-ho 4.OA.A.3 After 11:15 AM on Feb 26, list every chime through the end of that day. Half-hou 4.OA.A.3 The first $91$ chimes are used up on Feb 26. The chimes still needed to reach $2 4.NBT.B.6 Divide the remaining chimes by the daily total: $1912 \div 180 = 10$ with remain 4.OA.A.3 The $10$ full days start on Feb 27. Counting: Feb 27, Feb 28 (2 days), then Marc Review
Reasonableness: Sanity-check the size: $2003$ chimes at about $180$ per day is roughly $11$ days, and 11 days after Feb 26 lands in early-to-mid March — consistent with the choices near March 8 to March 10 and ruling out the far-away March 20 and 21. The remainder $112$ is well inside a single day's $180$, so the answer sits on the 11th day, not spilling into a 12th, which pins it to March 9.
Alternative: Instead of peeling off the partial first day, count total chimes from the start of Feb 26 (midnight): the whole day before 11:15 AM adds the midnight-through-11 AM chimes, and you can accumulate day by day, subtracting until you pass 2003. Either bookkeeping lands on 112 chimes into March 9, giving (B); batching by the 180-per-day rate is just the faster route to the same date.
CCSS standards used (min grade 4)
4.NBT.B.5Multiply a whole number of up to four digits by a one-digit whole number (Building the daily chime total $24 + 2\times 78 = 180$ from the half-hour and on-the-hour parts.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Tallying the 91 chimes left on Feb 26, subtracting to get 1912 remaining, and counting the days forward to the calendar date.)4.NBT.B.6Find whole-number quotients and remainders with up to four-digit dividends (Dividing $1912 \div 180 = 10$ remainder $112$ and reading the nonzero remainder as one extra partial day.)
⭐ When a count is huge, find how much one full day is worth, divide to get whole days plus a leftover, and the leftover tells you the extra day to land on.
⭐ When a count is huge, find how much one full day is worth, divide to get whole days plus a leftover, and the leftover tells you the extra day to land on.
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