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2024 AMC 10B Problem 4

2024 AMC 10B Problem 4AMC 10 Step by Step

Name: Date: October 28, 2026

Sequences & Series · difficulty 2 of 5 · about 3 min · position P1-10

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

A) $A$
B) $B$
C) $C$
D) $D$
E) $E$

Answer

Key insight

Step 63 ends at ball 2016, so ball 2024 is in step 64. Bins repeat every 5 steps, and step 64 lands on bin D.

Solution

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

Which step?

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:

step	1	2	3	4	5
last ball	1	3	6	10	15

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?

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:

remainder	1	2	3	4	0
bin	A	B	C	D	E

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

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

Why this works

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.

The trap

Finding step 64 correctly but then matching it to the wrong bin, for example calling it bin E. Check your rule against the example that step 7 is bin B.

Concise solution

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 $\boxed{\textbf{(D)}\ D}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 21: Sequences & Series (§21.1, §21.2, §21.3, §21.4)
Introduction to Number Theory, Chapter 12: Introduction to Modular Arithmetic
Intermediate Algebra, Chapter 17: More Sequences and Series (§17.1, §17.2)
Introduction to Number Theory, Chapter 10: Units Digits

Techniques

Bound the quantity above/below or estimate to pin it down · Set up the equation/formula and compute; no special trick needed