AMC 10 · 2024 · #4
Grade 6 arithmeticProblem
Balls numbered 1, 2, 3, ... are deposited in 5 bins, labeled A, B, C, D, and E, using the following procedure. Ball 1 is deposited in bin A, and balls 2 and 3 are deposited in bin B. The next 3 balls are deposited in bin C, the next 4 in bin D, and so on, cycling back to bin A after balls are deposited in bin E. (For example, balls numbered 22, 23, ..., 28 are deposited in bin B at step 7 of this process.) In which bin is ball 2024 deposited?
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: Balls numbered $1, 2, 3, \ldots$ are placed into bins $A, B, C, D, E$ in cycling order. Step $k$ places $k$ balls into one bin; step $1$ uses $A$, step $2$ uses $B$, $\ldots$, step $5$ uses $E$, step $6$ uses $A$ again, and so on. Which bin holds ball number $2024$?
Givens: Step $k$ deposits exactly $k$ balls; Bin order is $A, B, C, D, E$ cycling: step $k$ uses the bin at position $((k-1) \bmod 5) + 1$; Total balls placed through step $k$ equal $T_k = 1 + 2 + \cdots + k = \tfrac{k(k+1)}{2}$; Answer choices: (A) $A$, (B) $B$, (C) $C$, (D) $D$, (E) $E$
Unknowns: Which bin (A-E) contains ball $2024$
Understand
Restated: Balls numbered $1, 2, 3, \ldots$ are placed into bins $A, B, C, D, E$ in cycling order. Step $k$ places $k$ balls into one bin; step $1$ uses $A$, step $2$ uses $B$, $\ldots$, step $5$ uses $E$, step $6$ uses $A$ again, and so on. Which bin holds ball number $2024$?
Givens: Step $k$ deposits exactly $k$ balls; Bin order is $A, B, C, D, E$ cycling: step $k$ uses the bin at position $((k-1) \bmod 5) + 1$; Total balls placed through step $k$ equal $T_k = 1 + 2 + \cdots + k = \tfrac{k(k+1)}{2}$; Answer choices: (A) $A$, (B) $B$, (C) $C$, (D) $D$, (E) $E$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #6 Guess and Check, #9 Solve an Easier Related Problem
There are two patterns sitting on top of each other. Pattern #1: the cumulative ball count after step $k$ is the triangular number $T_k = \tfrac{k(k+1)}{2}$ — Tool #5 (Look for a Pattern) spots this by running steps $1, 2, 3$ and noticing the sums $1, 3, 6, 10, \ldots$. Pattern #2: the bin used at step $k$ depends only on $k \bmod 5$. Tool #6 (Guess and Check) is the fastest way to pin down which $k$ catches ball $2024$ — try $k$ near $\sqrt{2 \cdot 2024} \approx 64$. Tool #9 (Easier Related Problem) gives us the triangular-number formula itself by working out the small-$k$ cases.
Execute — Answer: D
6.EE.A.2 Step 1 - Find the cumulative count rule.
- After step $1$ there is $1$ ball; after step $2$ there are $1+2 = 3$; after step $3$, $1+2+3 = 6$; after step $4$, $10$.
- These are triangular numbers $T_k = \tfrac{k(k+1)}{2}$.
💡 Working out $k = 1, 2, 3, 4$ by hand reveals the closed form — a Grade 6 "write expressions with letters" move.
5.NBT.B.5 Step 2 - Estimate which step contains ball $2024$.
- Set $\tfrac{k(k+1)}{2} \approx 2024$, so $k(k+1) \approx 4048$ and $k \approx \sqrt{4048} \approx 63.6$.
- Check the two candidates near it.
💡 Squaring nearby values ($63^2 = 3969$, $64^2 = 4096$) brackets $k$ between $63$ and $64$ — Grade 5 multi-digit multiplication is enough to estimate.
5.NBT.B.5 Step 3 - Verify with $k = 63$ and $k = 64$.
- $T_{63} = \tfrac{63 \cdot 64}{2} = 63 \cdot 32 = 2016$, so balls $1$ through $2016$ are placed by the end of step $63$.
- Then step $64$ deposits balls $2017$ through $T_{64} = \tfrac{64 \cdot 65}{2} = 2080$.
💡 Two multi-digit products pin down which step ball $2024$ belongs to without a calculator — the Grade 5 multiplication skill, applied twice.
4.OA.B.4 Step 4 - Find the bin for step $64$ using the $5$-bin cycle.
- Compute $64 \bmod 5$.
- Since $64 = 12 \cdot 5 + 4$, the remainder is $4$.
💡 Cycling through $5$ bins means the bin only depends on the step number's remainder when divided by $5$ — a Grade 4 "factors and multiples" computation.
4.OA.C.5 Step 5 - Translate the remainder into a bin label.
- Remainder $1$ is $A$, $2$ is $B$, $3$ is $C$, $4$ is $D$, $0$ (or $5$) is $E$.
- So step $64$ uses bin $D$, and ball $2024$ lands there.
💡 Reading off the bin from a repeating $A, B, C, D, E$ pattern is the Grade 4 "generate a number or shape pattern from a rule" idea.
6.EE.A.2 Find the cumulative count rule. After step $1$ there is $1$ ball; after step $2$ 5.NBT.B.5 Estimate which step contains ball $2024$. Set $\tfrac{k(k+1)}{2} \approx 2024$, 5.NBT.B.5 Verify with $k = 63$ and $k = 64$. $T_{63} = \tfrac{63 \cdot 64}{2} = 63 \cdot 3 4.OA.B.4 Find the bin for step $64$ using the $5$-bin cycle. Compute $64 \bmod 5$. Since 4.OA.C.5 Translate the remainder into a bin label. Remainder $1$ is $A$, $2$ is $B$, $3$ Review
Reasonableness: Cross-check the cycle on a known case. The problem's example says balls $22, 23, \ldots, 28$ go into bin $B$ at step $7$. Step $7$ should use bin $B$ because $7 \equiv 2 \pmod 5$ — yes, matches. And $T_6 = 21$, $T_7 = 28$, so step $7$ deposits balls $22$ through $28$ — also matches. The same logic applied to $T_{64} = 2080$ and $64 \bmod 5 = 4$ gives bin $D$ for ball $2024$, confirming (D).
Alternative: Tool #2 (Make a Systematic List): list a few triangular-number checkpoints near $2024$. $T_{60} = 1830, T_{61} = 1891, T_{62} = 1953, T_{63} = 2016, T_{64} = 2080$. Ball $2024$ jumps over $T_{63}$ and lands inside step $64$. Then $64 \bmod 5 = 4$, so bin $D$ — the same answer (D).
CCSS standards used (min grade 6)
4.OA.B.4Find all factor pairs for a whole number in the range 1-100, and recognize multiples (Computing $64 \bmod 5$ by recognizing $64 = 12 \cdot 5 + 4$ to see which spot of the $5$-bin cycle is hit.)4.OA.C.5Generate a number or shape pattern following a given rule (Reading off bin $D$ from the repeating $A, B, C, D, E$ cycle once the step-number remainder is known.)5.NBT.B.5Fluently multiply multi-digit whole numbers (Computing $T_{63} = 63 \cdot 32 = 2016$ and $T_{64} = 32 \cdot 65 = 2080$ to bracket ball $2024$ between steps $63$ and $64$.)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Writing $T_k = \tfrac{k(k+1)}{2}$ as a closed-form expression and evaluating it at $k = 63$ and $k = 64$.)
⭐ This AMC 10 problem only needs Grade 6 expressions with letters (the formula $T_k = \tfrac{k(k+1)}{2}$) and Grade 4 remainders to map step $64$ to bin $D$!
⭐ This AMC 10 problem only needs Grade 6 expressions with letters (the formula $T_k = \tfrac{k(k+1)}{2}$) and Grade 4 remainders to map step $64$ to bin $D$!
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