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?

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

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

See also

2005 AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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