2005 AMC 10B Problem 11

Question

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

Compute the first several terms until the values start repeating

Once you find the repeating cycle, reduce the number of steps after the first term modulo its length

The sequence begins

2005, 133, 55, 250, 133, \ldots,

so after the first term it repeats the cycle

133, 55, 250

of length

3.

Terms

2,

3,

and

4

are the first, second, and third entries of this cycle. Because

2005 - 2 = 2003

leaves remainder

2

upon division by

3,

the

2005

th term matches the third entry,

250.

Thus, E is the correct answer.

Problem 11 in other years