AMC 10 · 2005 · #11
Grade 6 arithmeticThe first term of a sequence is 2005. Each succeeding term is the sum of the cubes of the digits of the previous term. What is the 2005th term of the sequence?
Pick an answer.
AMC 10 2005 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: A sequence starts at $2005$. To get each next term, take the previous term, cube each of its digits, and add those cubes together. Find the $2005^{\text{th}}$ term of the sequence.
Givens: The first term is $2005$; Each next term = sum of the cubes of the digits of the current term (for example, from $2005$ you compute $2^3+0^3+0^3+5^3$); Answer choices: (A) $29$, (B) $55$, (C) $85$, (D) $133$, (E) $250$
Unknowns: The value of the $2005^{\text{th}}$ term
Understand
Restated: A sequence starts at $2005$. To get each next term, take the previous term, cube each of its digits, and add those cubes together. Find the $2005^{\text{th}}$ term of the sequence.
Givens: The first term is $2005$; Each next term = sum of the cubes of the digits of the current term (for example, from $2005$ you compute $2^3+0^3+0^3+5^3$); Answer choices: (A) $29$, (B) $55$, (C) $85$, (D) $133$, (E) $250$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #7 Identify Subproblems, #9 Solve an Easier Related Problem
Computing $2005$ terms directly is impossible, so Tool #5 (Look for a Pattern) is the main move: generate the first several terms and watch for a value that repeats, which forces the whole sequence into a cycle. Tool #7 (Identify Subproblems) handles each single step — one cube-of-digits calculation at a time — so the arithmetic stays clean. Tool #9 (Solve an Easier Related Problem) is what finishes it: instead of reaching term $2005$ the long way, we replace that with an easy division-remainder question about where $2005$ lands inside the short repeating cycle.
Execute — Answer: E
6.EE.A.1 Step 1 Compute the second term
- Start from $2005$ and cube each digit, then add.
- The digits are $2$, $0$, $0$, $5$, so the second term is $2^3+0^3+0^3+5^3 = 8+0+0+125 = 133$.
💡 Cubing a digit just means multiplying it by itself three times, and the zeros drop out because $0^3=0$.
6.EE.A.1 Step 2 Compute the next few terms
- Keep applying the rule to each new term.
- From $133$: $1^3+3^3+3^3 = 1+27+27 = 55$.
- From $55$: $5^3+5^3 = 125+125 = 250$.
- From $250$: $2^3+5^3+0^3 = 8+125+0 = 133$.
💡 Each arrow is the same rule applied once, so you only ever need one small cube-and-add calculation at a time.
4.OA.C.5 Step 3 Spot the repeating cycle
- The fifth term came back to $133$, which was also the second term.
- Since each term depends only on the term right before it, once a value repeats the whole sequence must repeat forever.
- So from the second term onward the sequence cycles through three values: $133,\ 55,\ 250,\ 133,\ 55,\ 250,\dots$ with period $3$.
💡 A rule that only looks at the previous term must loop forever the moment any value shows up a second time.
4.NBT.B.6 Step 4 Find where the 2005th term lands
- The cycle covers terms $2,3,4,\dots$ with period $3$.
- Terms whose number leaves the same remainder when divided by $3$ share the same value: term $2$ (remainder $2$) $=133$, term $3$ (remainder $0$) $=55$, term $4$ (remainder $1$) $=250$.
- Now check the target: $2005 = 3\times 668 + 1$, so $2005$ leaves remainder $1$.
- That matches term $4$, whose value is $250$.
- So the $2005^{\text{th}}$ term is $250$, which is choice (E).
💡 In a length-$3$ loop, only the remainder after dividing by $3$ decides which of the three values you land on.
6.EE.A.1 Start from $2005$ and cube each digit, then add. The digits are $2$, $0$, $0$, $ 6.EE.A.1 Keep applying the rule to each new term. From $133$: $1^3+3^3+3^3 = 1+27+27 = 55 4.OA.C.5 The fifth term came back to $133$, which was also the second term. Since each te 4.NBT.B.6 The cycle covers terms $2,3,4,\dots$ with period $3$. Terms whose number leaves Review
Reasonableness: The answer must be one of the three cycle values $\{133, 55, 250\}$, and $250$ is on that list — a good sign. Choices (A) $29$ and (C) $85$ never appear in the sequence at all, so they are pure distractors. Choice (D) $133$ is the trap for an off-by-one slip: term $4$ is $250$ and matches remainder $1$, and $2005$ also has remainder $1$, so lining the target up with term $4$ (not term $2$) is the key. Since $2005 \equiv 1 \pmod 3$ exactly like term $4$, the value $250$ is right.
Alternative: Index the cycle from the second term instead of using remainders directly: the $k$-th cycle position (starting at term $2$ as position $0$) is $(n-2) \bmod 3$, giving $0\to133,\ 1\to55,\ 2\to250$. For $n=2005$: $(2005-2)=2003$, and $2003 = 3\times 667 + 2$, so the position is $2$, which is $250$. Same answer (E).
CCSS standards used (min grade 6)
6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Cubing each digit and summing, e.g. $2^3+0^3+0^3+5^3=133$ and $5^3+5^3=250$, to generate each successive term.)4.OA.C.5Generate a number or shape pattern following a given rule (Generating the terms from the given rule and recognizing that the values $133,55,250$ repeat with period $3$.)4.NBT.B.6Find whole-number quotients and remainders with up to four-digit dividends (Dividing $2005$ by $3$ to get remainder $1$, which locates the $2005^{\text{th}}$ term inside the length-$3$ cycle.)
⭐ When a rule keeps feeding a term back into itself, list a few terms until a value repeats, then use the remainder after dividing by the loop length to jump straight to any far-off term.
⭐ When a rule keeps feeding a term back into itself, list a few terms until a value repeats, then use the remainder after dividing by the loop length to jump straight to any far-off term.
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