AMC 10 · 2024 · #16
Grade 4 rate-ratioPick an answer.
AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The board changes in a structured way every move, so the work splits into two patterns to find — Tool #5. Pattern 1: the total count drops by 3 each move, so the final count is fixed mod 3. Pattern 2: count the odd numbers on the board; check how many odds disappear or appear per move. Both patterns are invariants (one strict, one a monovariant), and together they pin down the answer without any algebra. Tool #7 (Identify Subproblems) splits the work into (a) modular ceiling, (b) odd-count ceiling, (c) a constructive recipe showing 1010 is reachable. Tool #3 (Eliminate Possibilities) sweeps the five choices using the two ceilings — that is the most elementary AMC move and finishes the problem.
Split the start: 1 to 2024 gives 1012 evens and 1012 odds, 2024 numbers in all.
Splitting the starting list into odd and even buckets is the Grade 2 odd/even idea, just on a longer list.
2.OA.C.3Identify SubproblemsPattern 1: each move nets -3 to the count, so after k moves it is 2024 - 3k, forcing N ≡ 2 (mod 3).
When every step changes a quantity by the same fixed amount, the leftover after dividing by that amount cannot change — a Grade 4 "continue the pattern" rule.
4.OA.C.5Look For A PatternTest the choices mod 3: only 1010 and 1013 leave remainder 2, so B, C, E are out.
Divisibility-by-3 tests on a Grade 4 multiples chart instantly throw out three of the five choices.
4.OA.B.4Eliminate PossibilitiesPattern 2: track odd-count D. The three odd-producing moves change D by -3, 0, or -2 — so it never grows.
Listing the three odd-producing cases is plain Grade 2 odd-plus-even bookkeeping; the punchline is that none of them grows the odd count.
2.OA.C.3Look For A PatternD starts at 1012 and never grows, and the board ends all odd, so the final count is at most 1012 — kills 1013, leaving 1010.
Comparing the final count to the starting odd count is a Grade 1 two-digit-number comparison.
1.NBT.B.3Eliminate PossibilitiesPhase A: sum three evens and one odd (odd result) 337 times — clears 1011 evens, holds odds at 1012, leaving 1 even.
Repeated subtraction in groups of three is just Grade 3 division: 1012 ÷ 3 gives quotient 337 remainder 1.
3.OA.A.3Identify SubproblemsPhase B: sum that last even with three odds (odd) — odds fall to 1010, board all odd, so N = 1010, choice (A).
Putting the last piece in place is a Grade 2 take-away word problem: 1012 - 3 + 1 = 1010.
2.OA.A.1Identify SubproblemsThis AMC 10 problem only needs Grade 4 "continue-the-pattern" rules — count drops by 3 each move, odd count never grows — that you already know!