AMC 10 · 2003 · #22
Grade 4 arithmeticPick an answer.
AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Chimes in one full day
Half-hour chimes give 24 a day; hourly chimes give 1+2+…+12 = 78 twice, or 156. A full day holds 180 chimes.
Split the day into its steady half-hour ticks and its climbing hourly counts, then add the two totals.
4.NBT.B.5Analyze The UnitsChimes left on the first day
After 11:15 AM, list what is left: 13 half-hour chimes and 12 + 66 = 78 hourly ones, so Feb 26 still rings 91 chimes.
The starting time falls mid-day, so hand-count just the chimes left before the next midnight.
4.OA.A.3Make A Systematic ListHow many chimes remain
Feb 26 used 91 of them, so 2003 - 91 = 1912 chimes remain, and every one falls on Feb 27 or later at 180 per day.
Peel off the messy partial day first, so what is left is a whole number of clean full days plus a bit.
4.OA.A.3Identify SubproblemsDivide into whole days plus a remainder
Divide: 1912 ÷ 180 = 10 remainder 112. Ten whole days cover 1800, and the leftover 112 spills into one more day.
Ten whole days fit, and the leftover 112 chimes spill into one more day — that extra day is where the target lands.
Whole days fit a fixed number of times, and the leftover chimes spill into one more day.
▸ Why?
Dividing gives whole days plus a remainder smaller than a day, and that split is unique.
▸ Why?
Every full day chimes the same total, so the pattern repeats with that period.
Count the days forward
Feb 27, 28 plus March 1 through 8 are the 10 full days, so the leftover chimes ring on March 9, choice (B).
Ten days after Feb 27 ends on March 8, so the spillover chimes belong to March 9.
4.OA.A.3Make A Systematic ListWhen 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.
- Chimes in one full day
- Chimes left on the first day
- How many chimes remain
- Divide into whole days plus a remainder
- Count the days forward