2005 AMC 12B Problems/Problem 10

The following problem is from both the 2005 AMC 12B #10 and 2005 AMC 10B #11, so both problems redirect to this page.

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

$\textbf{(A) } 29 \qquad \textbf{(B) } 55 \qquad \textbf{(C) } 85 \qquad \textbf{(D) } 133 \qquad \textbf{(E) } 250$

Solution

Performing this operation several times yields the results of $133$ for the second term, $55$ for the third term, and $250$ for the fourth term. The sum of the cubes of the digits of $250$ equal $133$ , a complete cycle. The cycle is, excluding the first term, the $2^{\text{nd}}$ , $3^{\text{rd}}$ , and $4^{\text{th}}$ terms will equal $133$ , $55$ , and $250$ , following the fourth term. Any term number that is equivalent to $1\ (\text{mod}\ 3)$ will produce a result of $250$ . It just so happens that $2005\equiv 1\ (\text{mod}\ 3)$ , which leads us to the answer of $\boxed{\mathrm{(E)}\ 250}$ .

2005 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

2005 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2005 AMC 12B Problems/Problem 10

Problem 10

Solution

See also