Competition · AMC preparation · step 4 of 4

AMC 10 · 2024A · #10

Grade 6 number-theoryarithmetic
Archetype Modular Cycle Frequency Count └ Sub-archetype Position-in-Cycle Lookup
pattern-recognitionmodular-arithmeticrecursive-sequencedivisibility-rules pattern-recognitionsystematic-enumeration ↑ Prerequisites: divisibility-rulesmulti-digit-arithmetic
📏 Short solution 💡 2 insights
Problem
Start with n = 100. Repeat this rule: if n is a multiple of 3, replace it with n/3; otherwise replace it with n + 10. What is n after exactly 100 steps?

Pick an answer.

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

AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Look for a Pattern

A deterministic rule that runs 100 times on a small pool of positive integers is the classic setup for Tool #5 (Look for a Pattern): once a value repeats, the whole tail of the sequence becomes a cycle, and we never need to do all 100 steps. The natural way to expose that cycle is Tool #2 (Make an Organized List) — write the first several terms in order and watch for a value to come back. Once the cycle length is known, locating step 100 inside the cycle is just a remainder calculation.

1STEP 1

List the first terms

List the terms starting from 100, applying the rule each step; the digit-sum test makes divisibility by 3 quick.

100 → 110 → 120 → 40 → 50 → 60 → 20 → 30 → 10 → 20
2STEP 2

Spot the cycle

The value 20 returns after step 6 and step 9, so from step 6 on the sequence loops 20 → 30 → 10 in a cycle of length 3.

n₀, n₁, …, n₅_lead-in, n₆ = 20, n₇ = 30, n₈ = 10, n₉ = 20, …_cycle of length 3
3STEP 3

Locate step 100

From the cycle entry at step 6, the position at step 100 is (100 - 6) mod 3 = 94 mod 3 = 1.

100 - 6 = 94, 94 = 3 × 31 + 1 → 94 mod 3 = 1
4STEP 4

Read off the term

Position 0 is 20, position 1 is 30, position 2 is 10; step 100 sits at position 1, giving 30.

n₁₀₀ = cycle[1] = 30 → (C)
Answer
30
Double-check by counting a few steps forward from n₆ = 20 directly. n₇ = 30, n₈ = 10, n₉ = 20 (one full lap of length 3). So n₆ + 3k = 20 for every whole number k. Take k = 31: 6 + 3 × 31 = 99, so n₉₉ = 20. One more step: 20 is not a multiple of 3, so n₁₀₀ = 20 + 10 = 30. That matches answer (C) and the remainder argument. Also: the cycle values are {10, 20, 30}, all of which are answer choices, so the answer must be one of (A), (B), (C) — the structure of the choice list is itself a hint.
💡Key takeaway

Whenever a deterministic rule runs many times, list the first several terms and watch for a repeat — once you find the cycle, a single remainder calculation jumps you straight to step 100.

  • List the first terms
  • Spot the cycle
  • Locate step 100
  • Read off the term

A parent dashboard for the family lives at sensimlab.com.

More like this

Same problem type. Tags show what each one shares with this problem.