AMC 10 · 2013 · #13
Grade 4 countingPick an answer.
AMC 10 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The counting has a clean repeating rule: turn n contributes exactly n numbers. Spot that pattern, add the turn sizes to see how many numbers have been said by the end of each turn, then find which turn the 53rd number falls in and read off its value inside that turn.
Count numbers per turn
Jo says 1 number, Blair says 2, Jo says 3 — turn n says exactly n numbers, the list 1 through n.
The turn number is the same as how many numbers get said that turn.
The turn number is the same as how many numbers get said on that turn.
▸ Why?
Each turn says one more number than the last, so the counts climb by a fixed step.
▸ Why?
Each turn pairs with exactly one count, so the turn number names that count directly.
Add up to find the turn
Running totals: after turn 9, 1+2+…+9 = 45 numbers; after turn 10, 55. So 53 lands in turn 10.
Keep a running total of numbers said until it first reaches past 53.
4.NBT.B.4Identify SubproblemsLocate 53 inside turn 10
Turn 10 lists 1, 2, …, 10, and 53 - 45 = 8, so the 53rd number said is its 8th entry, 8.
Subtract the numbers already counted, then step that far into the current turn's list.
4.OA.A.3Make A Systematic ListAdd up how many numbers each turn says until you pass the one you want, then count that many steps into the last turn's list.
- Count numbers per turn
- Add up to find the turn
- Locate 53 inside turn 10
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