2024 AMC 10B Problem 4

There are two separate questions: which step puts ball $2024$ in a bin, and which bin that step uses.

Which step?

1. Step $1$ uses $1$ ball, step $2$ uses $2$ balls, step $3$ uses $3$ balls, and so on. So the last ball of each step is:

2. The last ball of step $n$ is $1 + 2 + \cdots + n = \frac{n(n+1)}{2}$.
3. We want this to be close to $2024$. Then $n(n+1)$ is close to $4048$, and $n \cdot n$ is about $4048$. Since $64 \cdot 64 = 4096$, try $n = 63$:
$$\frac{63 \cdot 64}{2} = 2016.$$
4. So step $63$ ends with ball $2016$. Step $64$ holds the next $64$ balls, $2017$ through $2080$, and ball $2024$ is among them.

Which bin?

1. Steps $1, 2, 3, 4, 5$ use bins $A, B, C, D, E$. Then steps $6, 7, 8, 9, 10$ use $A, B, C, D, E$ again, and so on. So only the remainder when the step number is divided by $5$ matters:

1. Check with the example: step $7$ leaves remainder $2$, which is bin $B$, just as the problem says.
2. Step $64$: $64 = 5 \cdot 12 + 4$, remainder $4$, so bin $D$.

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

When a process adds $1, 2, 3, \ldots$ items per step, the running totals are the triangular numbers $1, 3, 6, 10, \ldots$. To find where a large number falls, estimate with a square root and then check the exact totals on each side. The bin is a separate question that repeats every $5$ steps, so it depends only on the remainder. Always test your rule against the example the problem gives you.