2005 AMC 12B Problem 10

Below is the professionally curated solution for Problem 10 of the 2005 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2005 AMC 12B solutions, or check the answer key.

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Concepts:digits recursion pattern recognition

Difficulty rating: 1440

10.

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 $2005$ th term of the sequence?

$29$

$55$

$85$

$133$

$250$

Solution:

The sequence begins $2005, 133, 55, 250, 133, \ldots$ since $2^3 + 0^3 + 0^3 + 5^3 = 133,$ $1^3 + 3^3 + 3^3 = 55,$ $5^3 + 5^3 = 250,$ and $2^3 + 5^3 + 0^3 = 133.$

After the initial $2005,$ the terms cycle through $133, 55, 250$ with period $3.$

Term $n$ for $n \ge 2$ is the $((n-2)\bmod 3)$ th entry of $133, 55, 250.$ Since $2005 - 2 = 2003 \equiv 2 \pmod 3,$ the $2005$ th term is $250.$

Thus, the correct answer is E.

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