2004 AMC 10B Problem 19

Attempt Problem 19 of the 2004 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2004 AMC 10B solutions, or check the answer key.

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19.

In the sequence 2001,2002,2003,,2001, 2002, 2003, \ldots, each term after the third is found by subtracting the previous term from the sum of the two terms that precede that term. For example, the fourth term is 2001+20022003=2000.2001 + 2002 - 2003 = 2000. What is the 20042004th term in this sequence?

2004-2004

2-2

00

40034003

60076007

Answer: C
Concepts:recursionarithmetic sequencepattern recognition
Difficulty rating: 1460
Solution:

The recurrence ak+1=ak2+ak1aka_{k+1} = a_{k-2} + a_{k-1} - a_k gives ak+1ak1=(akak2).a_{k+1} - a_{k-1} = -(a_k - a_{k-2}). The sequence begins 2001,2002,2003,2001, 2002, 2003, 2000,2005,1998,2000, 2005, 1998, \ldots

So the even-position terms form the arithmetic sequence 2002,2000,1998,2002, 2000, 1998, \ldots with common difference 2.-2. The 20042004th term is its 10021002nd term, 2002+1001(2)=0.2002 + 1001(-2) = 0.

Thus, the correct answer is C.

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