AMC 10 · 2004 · #1
Grade 4 arithmeticPick an answer.
AMC 10 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The problem splits into two clean subproblems (Tool #7): first count how many rows run from 12 to 22, then multiply that count by the 33 seats in each row. The one real trap is the inclusive count — writing 22-12=10 forgets that row 12 itself is reserved. Tool #3 (Eliminate Possibilities) shows every wrong choice is a specific miscount: 330=10×33 (off by one), 297=9×33, 396=12×33, and 726=22×33.
Count the reserved rows
Both endpoints count, so subtract and add one: 22-12+1 = 11 reserved rows, not 10.
Counting a run of numbers that includes both ends means subtract the ends, then add one for the starting row.
Counting a run of numbers that includes both ends means subtracting the ends and adding one back.
▸ Why?
Each row in the block is one item and each gap is a space between two of them, so items outnumber gaps by one.
▸ Why?
The count of gaps is just the difference of the endpoints, wherever the block sits.
Multiply rows by seats per row
The 11 rows are equal groups of 33 seats, so the total is 11×33 = 330 + 33 = 363.
Equal groups multiply: number of rows times seats in each row gives the total seats.
4.NBT.B.5Identify SubproblemsCheck against the trap choices
363 is choice (C); the traps count 10 rows for (B) 330, 9 for (A) 297, 12 for (D) 396, and all 22 for (E) 726.
Each wrong choice is a specific off-by-one or over-count, so nailing the row count pins the answer.
4.OA.A.3Eliminate PossibilitiesTo count rows from a first number through a last number, subtract them and add one — then multiply by how many seats each row holds.
- Count the reserved rows
- Multiply rows by seats per row
- Check against the trap choices