Question

Balls numbered $1, 2, 3, \ldots$ are placed in bins $A, B, C, D,$ and $E$ so that the first ball is placed in $A,$ the next two are placed in $B,$ the next three are placed in $C,$ the next four are placed in $D,$ the next five are placed in $E,$ and then the next six go in $A,$ etc. For example, $22, 23, \ldots, 28$ are placed in $B.$ Which bin contains ball $2024?$

$A$

$B$

$C$

$D$

$E$

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

Solution:

Problem 4 in Other Years