AMC 10 · 2024 · #4

Grade 6 arithmetic
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)
A
(B)
B
(C)
C
(D)
D
(E)
E
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 = (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

Write the running total

The total after k steps is a triangular number.

T_k = 1 + 2 + … + k = (k(k+1))/2
2STEP 2

Estimate the step

Around 2024 the step is 63 or 64.

k(k+1) ≈ 4048 → k ≈ 63 or 64
3STEP 3

Pin the exact step

Ball 2024 lands in step 64.

T₆₃ = 2016, T₆₄ = 2080, 2017 ≤ 2024 ≤ 2080
4STEP 4

Take it mod five

64 leaves remainder 4.

64 = 12 · 5 + 4 → 64 ≡ 4 (mod 5)
5STEP 5

Read the bin

The fourth bin is D.

64 ≡ 4 (mod 5) ⟹ bin D → (D)
Answer
D
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 12 problem only needs Grade 6 expressions with letters (the formula T_k = (k(k+1))/2) and Grade 4 remainders to map step 64 to bin D!