2005 AMC 10B Problem 11

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 $\boxed{\textbf{(E)}\ 250}$.

A digit-based recursion on small numbers must eventually revisit a value, after which it is periodic. Compute until the first repeat, note where the cycle starts (here term 2, not term 1), and reduce the target index modulo the period relative to that starting point.

• Aligning the cycle with term 1 instead of term 2, which shifts the answer to 133 or 55.
• Reducing $2005 \bmod 3$ correctly but mapping remainder $1$ to the first cycle element ($133$) rather than to the slot occupied by term $4$.