AMC 10 · 2024 · #16
Grade 4 rate-ratioProblem
Jerry likes to play with numbers. One day, he wrote all the integers from 1 to 2024 on the whiteboard. Then he repeatedly chose four numbers on the whiteboard, erased them, and replaced them by either their sum or their product. (For example, Jerry's first step might have been to erase 1, 2, 3, and 5, and then write either 11, their sum, or 30, their product, on the whiteboard.) After repeatedly performing this operation, Jerry noticed that all the remaining numbers on the whiteboard were odd. What is the maximum possible number of integers on the whiteboard at that time?
Pick an answer.
AMC 10 2024 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: The integers from $1$ to $2024$ are on a whiteboard. Each move: pick any $4$ numbers, erase them, and write either their sum or their product. After some sequence of moves, every remaining number is odd. What is the largest possible count of numbers still on the board?
Givens: Start: integers $1, 2, 3, \dots, 2024$ — so $1012$ even and $1012$ odd; Each move replaces $4$ numbers by $1$ number, so total count drops by $3$; The replacement value is the sum OR the product of the chosen $4$; Final state: every number on the board is odd; Answer choices: (A) $1010$, (B) $1011$, (C) $1012$, (D) $1013$, (E) $1014$
Unknowns: The maximum possible number of integers remaining when the board is all odd
Understand
Restated: The integers from $1$ to $2024$ are on a whiteboard. Each move: pick any $4$ numbers, erase them, and write either their sum or their product. After some sequence of moves, every remaining number is odd. What is the largest possible count of numbers still on the board?
Givens: Start: integers $1, 2, 3, \dots, 2024$ — so $1012$ even and $1012$ odd; Each move replaces $4$ numbers by $1$ number, so total count drops by $3$; The replacement value is the sum OR the product of the chosen $4$; Final state: every number on the board is odd; Answer choices: (A) $1010$, (B) $1011$, (C) $1012$, (D) $1013$, (E) $1014$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities
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.
Execute — Answer: A
2.OA.C.3 Step 1 - Count parity at the start.
- Among $1, 2, \dots, 2024$ there are $2024 / 2 = 1012$ even numbers and $1012$ odd numbers, for a total of $2024$.
💡 Splitting the starting list into odd and even buckets is the Grade 2 odd/even idea, just on a longer list.
4.OA.C.5 Step 2 - Pattern 1: total count drops by $3$ each move.
- Each move erases $4$ numbers and writes $1$, so the net change is $-3$.
- After $k$ moves the board has $2024 - 3k$ numbers, so the final count $N$ satisfies $N \equiv 2024 \equiv 2 \pmod 3$.
💡 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.B.4 Step 3 - Use Pattern 1 to eliminate choices.
- Check $1010, 1011, 1012, 1013, 1014$ modulo $3$: only $1010 \equiv 2$ and $1013 \equiv 2$ survive.
💡 Divisibility-by-$3$ tests on a Grade 4 multiples chart instantly throw out three of the five choices.
2.OA.C.3 Step 4 - Pattern 2: track how the odd-count $D$ changes per move.
- The move's output is odd only in three cases — list them: (i) Product of four odds gives one odd (uses $4$ odds, gains $1$): $D$ drops by $3$.
- (ii) Sum of three evens and one odd gives one odd (uses $1$ odd, gains $1$): $D$ stays the same.
- (iii) Sum of three odds and one even gives one odd (uses $3$ odds, gains $1$): $D$ drops by $2$.
- In every move that ends with an odd written down, the odd-count never increases.
💡 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.
1.NBT.B.3 Step 5 - Conclude the second ceiling.
- Since the odd count $D$ starts at $1012$ and never grows, and since the final board is all odd, the final count equals $D_{\text{final}} \le 1012$.
- So $1013$ is impossible, eliminating choice (D).
- The only survivor is choice (A) $1010$.
💡 Comparing the final count to the starting odd count is a Grade 1 two-digit-number comparison.
3.OA.A.3 Step 6 - Constructive check that $1010$ is actually reachable.
- Phase A — kill evens cheaply.
- Use case (iii) of step 4: pick three evens and one odd, take their SUM (odd).
- Run this $337$ times: each run kills $3$ evens and keeps the odd-count steady at $1012$ (uses $1$ odd, makes $1$ odd).
- After $337$ runs, $337 \cdot 3 = 1011$ evens are gone, leaving $1$ even and still $1012$ odds.
💡 Repeated subtraction in groups of three is just Grade 3 division: $1012 \div 3$ gives quotient $337$ remainder $1$.
2.OA.A.1 Step 7 - Phase B — kill the last even.
- Apply case (iii) once more: take the remaining $1$ even and any $3$ odds and SUM them (odd).
- That move uses $3$ odds and makes $1$, so the odd count drops by $2$ to $1010$.
- The board is now all odd with $1010$ numbers.
💡 Putting the last piece in place is a Grade 2 take-away word problem: $1012 - 3 + 1 = 1010$.
2.OA.C.3 Count parity at the start. Among $1, 2, \dots, 2024$ there are $2024 / 2 = 1012$ 4.OA.C.5 Pattern 1: total count drops by $3$ each move. Each move erases $4$ numbers and 4.OA.B.4 Use Pattern 1 to eliminate choices. Check $1010, 1011, 1012, 1013, 1014$ modulo 2.OA.C.3 Pattern 2: track how the odd-count $D$ changes per move. The move's output is od 1.NBT.B.3 Conclude the second ceiling. Since the odd count $D$ starts at $1012$ and never 3.OA.A.3 Constructive check that $1010$ is actually reachable. Phase A — kill evens cheap 2.OA.A.1 Phase B — kill the last even. Apply case (iii) once more: take the remaining $1$ Review
Reasonableness: Two independent ceilings — "count $\equiv 2 \pmod 3$" and "final count $\le$ starting odd count $= 1012$" — leave only $1010$ among the choices, and the explicit recipe lands on exactly $1010$. Sanity: total moves used $= 338$, total count drop $= 338 \times 3 = 1014$, and $2024 - 1014 = 1010$, matching $N$. Also $1010 \equiv 2 \pmod 3$, consistent with Pattern 1. Both ceilings are met with equality, so (A) is both an upper bound and achievable.
Alternative: Tool #3 (Eliminate Possibilities) standing alone: just test the five choices against the mod-$3$ rule of Pattern 1 — that already discards (B), (C), (E). Then a one-line check of Pattern 2 ("odd count never grows") rules out (D). No construction is needed if you trust the existence claim, but the recipe above makes the achievability concrete.
CCSS standards used (min grade 4)
2.OA.C.3Determine whether a group of objects has an odd or even number (Counting that the starting board has $1012$ evens and $1012$ odds, and tracking the parity of sums and products of four numbers.)4.OA.C.5Generate a number or shape pattern following a given rule (Recognizing that the total board count drops by exactly $3$ each move, so the final count is congruent to $2024 \equiv 2 \pmod 3$.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Testing each of the five answer choices against divisibility by $3$ to keep only $1010$ and $1013$.)1.NBT.B.3Compare two two-digit numbers using symbols (Comparing the final count to the starting odd count $1012$ to rule out $1013$.)3.OA.A.3Solve multiplication and division word problems within 100 (Computing $1012 = 3 \times 337 + 1$ to plan how many EEEO-sum moves clear all but one even.)2.OA.A.1Solve one- and two-step word problems using addition and subtraction within 100 (Bookkeeping the final odd count: $1012 - 3 + 1 = 1010$ after the last EOOO-sum move.)
⭐ This 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!
⭐ This 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!
More like this
Same archetype — closest grade level first.
