2024 AMC 10B Problem 4

Step $k$ deposits $k$ balls, so after step $n$ a total of $1 + 2 + \cdots + n = \frac{n(n+1)}{2}$ balls have been placed. The bin for step $k$ depends only on $k \bmod 5$: steps $1, 2, 3, 4, 5$ go to $A, B, C, D, E$, so remainder $1 \to A$, $2 \to B$, $3 \to C$, $4 \to D$, $0 \to E$. (Check: step $7$ has remainder $2$ and indeed uses bin $B$.)

We need the step containing ball $2024$. Since $\frac{n(n+1)}{2} \approx 2024$ means $n^2 \approx 4048$, try $n = 63$: $\frac{63 \cdot 64}{2} = 2016$. So step $63$ ends with ball $2016$, and step $64$ covers balls $2017$ through $2080$, which includes $2024$.

Finally $64 = 5 \cdot 12 + 4$, remainder $4$, which is bin $D$.

The answer is $\boxed{\textbf{(D)}\ D}$.

A process that deposits $1, 2, 3, \dots$ items per step is governed by triangular numbers; locate the step by solving $\frac{n(n+1)}{2} \approx N$ with a square-root estimate, then verify with the exact values on either side. The bin is a separate, purely periodic question, answered by the step number mod $5$. Always test the mapping against the example the problem supplies.

• Mapping the step number to the wrong bin, e.g. sending step 64 to bin E by treating remainder 4 as the fifth letter or starting the count at zero.
• Using $2024 \bmod 5$ (the ball number) instead of the step number, which gives remainder $4$ by coincidence here but is the wrong quantity.