2005 AMC 12B Problem 10

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 and look for a repeating cycle

Once a value repeats, the later terms repeat with the same period; use the term index modulo that period

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.

Problem 10 in other years