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2024 AMC 10A Problem 10

2024 AMC 10A Problem 10AMC 10 Step by Step

Name: Date: October 28, 2026

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

Consider the following operation. Given a positive integer $n$ , if $n$ is a multiple of $3$ , then you replace $n$ by $\frac{n}{3}$ . If $n$ is not a multiple of $3$ , then you replace $n$ by $n+10$ . For example, beginning with $n=4$ , this procedure gives $4\to14\to24\to8\to18\to6\to2\to12\to\cdots$ . Suppose you start with $n=100$ . What value results if you perform this operation exactly $100$ times?

A) $10$
B) $20$
C) $30$
D) $40$
E) $50$

Answer

Key insight

Work out the first ten or so steps. From step 6 the values repeat 20, 30, 10, so step 100 matches step 7, which is 30.

Solution

Doing $100$ steps by hand is too many, but the numbers might start repeating. So do the first few steps and watch.

Start at $100$ . It is not a multiple of $3$ , so add $10$ : $110$ . Keep going:

step	1	2	3	4	5	6	7	8	9	10	11
value	110	120	40	50	60	20	30	10	20	30	10

(For example, $120$ is a multiple of $3$ , so it becomes $40$ ; $60$ becomes $20$ ; $30$ becomes $10$ ; $10$ is not a multiple of $3$ , so it becomes $20$ .)
2. At step $9$ we are back at $20$ , the value from step $6$ . From then on the same three numbers repeat forever: $20, 30, 10, 20, 30, 10, \ldots$
3. Match each step with its value:
- steps $6, 9, 12, \ldots$ (multiples of $3$ ) give $20$ ,
- steps $7, 10, 13, \ldots$ (one more than a multiple of $3$ ) give $30$ ,
- steps $8, 11, 14, \ldots$ (two more than a multiple of $3$ ) give $10$ .
4. $100 = 99 + 1$ , and $99 = 3 \cdot 33$ . So $100$ is one more than a multiple of $3$ , just like step $7$ . The value is $30$ .

The answer is $\boxed{\textbf{(C)}\ 30}$ .

Why this works

A rule that always does the same thing to the same number must start repeating as soon as a number comes back. Work out steps until you see a repeat, note how long the repeating block is, and then use division to find where the target step lands in the block. Line it up using a step you have actually computed, so you do not end up one step off.

The trap

Lining the repeating pattern up with the wrong step. Step 6 is 20, so steps 7, 10, 13, and so on (one more than a multiple of 3) are 30.

Concise solution

Apply the rule and record the result after each step:

step	1	2	3	4	5	6	7	8	9	10	11
value	110	120	40	50	60	20	30	10	20	30	10

From step $6$ on the values repeat $20, 30, 10$ with period $3$ : step $6, 9, 12, \dots$ (multiples of $3$ ) give $20$ ; steps $7, 10, 13, \dots$ give $30$ ; steps $8, 11, 14, \dots$ give $10$ .

Since $100 = 3\cdot 33 + 1$ is one more than a multiple of $3$ (like step $7$ ), the value after $100$ operations is $30$ .

The answer is $\boxed{\textbf{(C)}\ 30}$ .

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
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.1)
Intermediate Algebra, Chapter 17: More Sequences and Series (§17.1, §17.2)

Techniques

Compute small cases, spot the pattern, generalize