AMC 10 · 2024 · #10
Grade 6 number-theoryarithmeticProblem
Consider the following operation. Given a positive integer n, if n is a multiple of 3, then you replace n by 3n. If n is not a multiple of 3, then you replace n by n+10. Then continue this process. For example, beginning with n=4, this procedure gives 4→14→24→8→18→6→2→12→⋯. Suppose you start with n=100. What value results if you perform this operation exactly 100 times?
Pick an answer.
AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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?
Givens: Starting value $n_0 = 100$; Rule: if $3 \mid n$, then $n \to n/3$; if $3 \nmid n$, then $n \to n + 10$; We perform the operation exactly $100$ times; Answer choices: (A) $10$, (B) $20$, (C) $30$, (D) $40$, (E) $50$
Unknowns: The value of $n$ after step $100$
Understand
Restated: 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?
Givens: Starting value $n_0 = 100$; Rule: if $3 \mid n$, then $n \to n/3$; if $3 \nmid n$, then $n \to n + 10$; We perform the operation exactly $100$ times; Answer choices: (A) $10$, (B) $20$, (C) $30$, (D) $40$, (E) $50$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #2 Make an Organized List
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.
Execute — Answer: C
4.OA.C.5 Step 1 - List the first several terms.
- Start at $100$, apply the rule each step, and write the result.
- Use the divisibility-by-$3$ shortcut: a number is a multiple of $3$ iff its digit sum is.
- ($100 \to 1$, not divisible; $120 \to 3$, divisible; $60 \to 6$, divisible; etc.)
💡 Generating each next term from the rule is a Grade 4 "follow a given rule to create a number pattern" exercise.
4.OA.C.5 Step 2 - Spot the cycle.
- The value $20$ shows up after step $6$ and again after step $9$, so between those two appearances the sequence travels $20 \to 30 \to 10 \to 20$.
- That is a cycle of length $3$.
- From step $6$ onward the sequence is locked into this loop forever.
💡 Once any value repeats in a deterministic process, the same chain of successors must repeat too — that is what makes the loop genuine.
6.NS.B.2 Step 3 - Locate step $100$ inside the cycle.
- The cycle entry is step $6$, so steps $6, 7, 8, 9, \ldots$ trace $20, 30, 10, 20, \ldots$ in lockstep with the cycle positions $0, 1, 2, 0, \ldots$.
- The cycle position at step $k$ (for $k \ge 6$) is $(k - 6) \bmod 3$.
💡 Dividing the offset $94$ by the cycle length $3$ and keeping the remainder tells you how many "extra" steps remain after completing whole laps — the standard skip-ahead trick for cycles.
4.OA.C.5 Step 4 - Read off the answer.
- Cycle position $0$ is $20$, position $1$ is $30$, position $2$ is $10$.
- Step $100$ sits at position $1$, so $n_{100} = 30$.
💡 The remainder is just an index into the three-element cycle list — no further arithmetic needed.
4.OA.C.5 List the first several terms. Start at $100$, apply the rule each step, and writ 4.OA.C.5 Spot the cycle. The value $20$ shows up after step $6$ and again after step $9$, 6.NS.B.2 Locate step $100$ inside the cycle. The cycle entry is step $6$, so steps $6, 7, 4.OA.C.5 Read off the answer. Cycle position $0$ is $20$, position $1$ is $30$, position Review
Reasonableness: Double-check by counting a few steps forward from $n_6 = 20$ directly. $n_7 = 30$, $n_8 = 10$, $n_9 = 20$ (one full lap of length $3$). So $n_{6 + 3k} = 20$ for every whole number $k$. Take $k = 31$: $6 + 3 \times 31 = 99$, so $n_{99} = 20$. One more step: $20$ is not a multiple of $3$, so $n_{100} = 20 + 10 = 30$. 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.
Alternative: Tool #6 (Guess and Check) directly on landmark steps: from step $6$ the sequence is $20, 30, 10, 20, 30, 10, \ldots$, so $n_k$ equals $20$ when $k \equiv 6 \pmod 3$ (i.e. $k$ is a multiple of $3$), equals $30$ when $k \equiv 1 \pmod 3$, and equals $10$ when $k \equiv 2 \pmod 3$. Since $100 \equiv 1 \pmod 3$, $n_{100} = 30$. Same answer (C) without computing the offset $94$ explicitly.
CCSS standards used (min grade 6)
4.OA.C.5Generate a number pattern that follows a given rule and identify apparent features of the pattern (Producing the first $9$ terms of the sequence from the rule and observing the repeating loop $20 \to 30 \to 10 \to 20$.)4.OA.B.4Find factor pairs and recognize multiples of small whole numbers (Deciding at each step whether the current value is a multiple of $3$ (using the digit-sum test) so we know which branch of the rule to apply.)6.NS.B.2Fluently divide multi-digit numbers using the standard algorithm (Computing $94 = 3 \times 31 + 1$ to locate step $100$ inside a cycle of length $3$.)
⭐ 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 $100$.
⭐ 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 $100$.
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