AMC 10 · 2024 · #4

Grade 6 arithmetic

Archetype Modular Cycle Frequency Count └ Sub-archetype Position-in-Cycle Lookup

sequences-arithmeticmodular-arithmeticpattern-recognition pattern-recognitionidentify-subproblemsconvert-to-algebra ↑ Prerequisites: sequences-arithmeticmodular-arithmeticmulti-digit-arithmetic

📏 Medium solution 💡 3 insights

Problem

Balls numbered 1, 2, 3, … 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, …, step 5 uses E, step 6 uses A again, and so on. Which bin holds ball number 2024?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Look for a Pattern

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 = $\frac{k(k+1)}{2}$ — Tool #5 (Look for a Pattern) spots this by running steps 1, 2, 3 and noticing the sums 1, 3, 6, 10, …. Pattern #2: the bin used at step k depends only on k mod 5. Tool #6 (Guess and Check) is the fastest way to pin down which k catches ball 2024 — try k near √(2 · 2024) ≈ 64. Tool #9 (Easier Related Problem) gives us the triangular-number formula itself by working out the small-k cases.

1STEP 1

Adding up 1, 2, 3, … by hand gives running totals 1, 3, 6, 10 — the triangular numbers T_k = $\frac{k(k+1)}{2}$ .

T_k = 1 + 2 + … + k =

\frac{k(k+1)}{2}

2STEP 2

Set $\frac{k(k+1)}{2}$ ≈ 2024, so k(k+1) ≈ 4048 and k ≈ √4048 ≈ 63.6 — so k is 63 or 64.

k(k+1) ≈ 4048 → k ≈ 63 or 64

3STEP 3

By step 63, 2016 balls are placed; step 64 covers balls 2017 through 2080, so 2024 falls in step 64.

T₆3 = 2016, T₆4 = 2080, 2017 ≤ 2024 ≤ 2080

4STEP 4

The bin depends on the step number mod 5. Since 64 = 12 · 5 + 4, we get 64 ≡ 4 (mod 5).

64 = 12 · 5 + 4 → 64 ≡ 4 (mod 5)

5STEP 5

Remainder 1 is A, 2 is B, 3 is C, 4 is D, 0 is E. Remainder 4 means step 64 uses bin D, so ball 2024 lands there.

64 ≡ 4 (mod 5) ⟹ bin D → (D)

Answer

Cross-check the cycle on a known case. The problem's example says balls 22, 23, …, 28 go into bin B at step 7. Step 7 should use bin B because 7 ≡ 2 (mod 5) — yes, matches. And T₆ = 21, T₇ = 28, so step 7 deposits balls 22 through 28 — also matches. The same logic applied to T₆4 = 2080 and 64 mod 5 = 4 gives bin D for ball 2024, confirming (D).

💡Key takeaway

This AMC 10 problem only needs Grade 6 expressions with letters (the formula T_k = $\frac{k(k+1)}{2}$ ) and Grade 4 remainders to map step 64 to bin D!

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