Question

Balls numbered $1, 2, 3, \ldots$ 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 $22, 23, \ldots, 28$ are deposited in bin $B$ at step $7$ of this process.) In which bin is ball $2024$ deposited?

$A$

$B$

$C$

$D$

$E$

Answer: D

Concepts:triangular number modular arithmetic

Difficulty rating: 1130

Solution:

Group $g$ holds $g$ balls, so the first $g$ groups swallow $\tfrac{g(g+1)}{2}$ of them. Now $\tfrac{63 \cdot 64}{2} = 2016$ and $\tfrac{64 \cdot 65}{2} = 2080,$ which puts ball $2024$ in group $64$ (balls $2017$ through $2080$ ). The bins cycle $A, B, C, D, E,$ so group $g$ lands in bin number $(g - 1) \bmod 5.$ For $g = 64$ that's $63 \bmod 5 = 3,$ bin $D.$ Therefore, the answer is D.

2024 AMC 10B Problem 4

Answer: D

Difficulty rating: 1130

Solution:

Problem 4 in Other Years