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2005 AMC 10B Problem 11

2005 AMC 10B Problem 11AMC 10 Step by Step

Name: Date: October 28, 2026

Sequences & Series · difficulty 3 of 5 · about 3 min · position P11-15

The 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 ${2005}^{\text{th}}$ term of the sequence?

A) $29$
B) $55$
C) $85$
D) $133$
E) $250$

Answer

Key insight

Write out terms until one repeats. From term 2 on, the sequence cycles 133, 55, 250 every 3 terms, and term 2005 lands on 250, like term 4.

Solution

Computing $2005$ terms by hand is impossible, so look for a repeat.

1. Write out terms.
- Term 1: $2005$ .
- Term 2: $2^3 + 0^3 + 0^3 + 5^3 = 8 + 125 = 133$ .
- Term 3: $1^3 + 3^3 + 3^3 = 1 + 27 + 27 = 55$ .
- Term 4: $5^3 + 5^3 = 125 + 125 = 250$ .
- Term 5: $2^3 + 5^3 + 0^3 = 8 + 125 = 133$ . This is term 2 again!
2. The pattern. Since each term depends only on the one before it, after $133$ we get $55$ , then $250$ , then $133$ again, forever. The cycle has length $3$ and starts at term 2:

Term	2	3	4	5	6	7	8	...
Value	133	55	250	133	55	250	133	...

Which slot is term 2005? The value $250$ appears at terms $4, 7, 10, \ldots$ . These are the numbers that leave remainder $1$ when divided by $3$ . Since $2005 = 3 \cdot 668 + 1$ , term $2005$ leaves remainder $1$ too, so it is $250$ .

The answer is $\boxed{\textbf{(E)}\ 250}$ .

Why this works

When each term is built from the digits of the one before, the numbers quickly get small and must eventually repeat. Once a value comes back, the sequence cycles. Be careful to note where the cycle begins (here at term 2), and match the target term to a known term with the same remainder.

The trap

Starting the cycle count at term 1 instead of term 2. Term 1 (2005) is not part of the cycle, and shifting the count gives 133 or 55.

Concise solution

Compute terms until something repeats.

Term 1: $2005$ .
Term 2: $2^3 + 0 + 0 + 5^3 = 8 + 125 = 133$ .
Term 3: $1 + 27 + 27 = 55$ .
Term 4: $125 + 125 = 250$ .
Term 5: $8 + 125 + 0 = 133$ , which equals term 2.

From term 2 onward the sequence repeats with period $3$ : terms $2, 5, 8, \dots$ are $133$ ; terms $3, 6, 9, \dots$ are $55$ ; terms $4, 7, 10, \dots$ are $250$ .

Term $n \ge 2$ is $250$ exactly when $n \equiv 1 \pmod 3$ . Since $2005 = 3 \cdot 668 + 1$ , term $2005$ is $250$ .

The answer is $\boxed{\textbf{(E)}\ 250}$ .

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 8: Base Numbers
Introduction to Number Theory, Chapter 12: Introduction to Modular Arithmetic
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.1)

Techniques

Compute small cases, spot the pattern, generalize