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2024 AMC 10A Problem 10

2024 AMC 10A Problem 10AMC 10 Step by Step

Name: Date: September 13, 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

Iterate: 100→110→120→40→50→60→20→30→10→20→…; from step 6 the values cycle 20, 30, 10 with period 3, and step 100 ≡ step 7 gives 30.

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 $\boxed{\textbf{(C)}\ 30}$ .

Why this works

A deterministic rule on a finite set of reachable values must eventually cycle; once a value repeats, everything after it is periodic. Compute until the first repeat, note where the cycle starts and its length, then reduce the target step number modulo the period, taking care to align with a known step inside the cycle.

The trap

Off-by-one when aligning the cycle: step 6 is 20, so steps 7, 10, 13, … (one more than a multiple of 3) are 30.

Common mistakes

Off-by-one when aligning the cycle: step 6 is 20, so steps 7, 10, 13, … (one more than a multiple of 3) are 30.
Assuming the cycle starts at step $1$ and reducing $100 \bmod 3$ against the wrong anchor.

Techniques

Compute small cases, spot the pattern, generalize