AMC 10 · 2004 · #19

Grade 6 arithmetic
Archetype Lattice / Grid Pattern from Small Cases └ Sub-archetype Periodic Placement Formula
recursive-sequencepattern-recognitionmodular-arithmetic pattern-recognitionsystematic-enumeration ↑ Prerequisites: recursive-sequence
📏 Medium solution 💡 3 insights
Problem
A sequence starts 2001, 2002, 2003, … From the fourth term on, each term is found by adding the two terms that come before the previous term, then subtracting the previous term. For instance, the fourth term is 2001 + 2002 - 2003 = 2000. What is the 2004th term of this sequence?

Pick an answer.

(A)
-2004
(B)
-2
(C)
0
(D)
4003
(E)
6007

AMC 10 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Look for a Pattern

Reaching term 2004 one step at a time is hopeless, so the move is to compute enough early terms to spot a pattern (Tool #5). The raw list 2001, 2002, 2003, 2000, 2005, 1998, 2007, 1996, … looks jumpy, but splitting it into odd-numbered and even-numbered positions (Tool #15) makes each half a clean, steady sequence. Since 2004 is even, only the even-position half matters. Finally, the answer choices are far apart, so a correct value pins one choice and rules out the rest (Tool #3).

1STEP 1

Generate the first several terms

Apply the rule repeatedly to get the next terms 2000, 2005, 1998, 2007, 1996 — the list zig-zags.

a₄=2000, a₅=2005, a₆=1998, a₇=2007, a₈=1996
2STEP 2

Split odd and even positions

Sort by position: the odd terms 2001, 2003, 2005, 2007 climb by 2, while the even terms 2002, 2000, 1998, 1996 fall by 2 each time.

even positions: 2002, 2000, 1998, 1996, … (down by 2)
3STEP 3

Extend the even pattern to position 2004

The kth even term is 2002 - 2(k-1), and position 2004 is the 1002nd even term, so it equals 2002 - 2·1001 = 0.

a₂₀₀₄ = E₁₀₀₂ = 2002 - 2 · 1001 = 0
4STEP 4

Match the value to a choice

The 2004th term is 0, which is choice (C); no other option fits the even-position pattern.

a₂₀₀₄ = 0 → (C)
Answer
0
The even terms 2002, 2000, 1998, … march steadily downward by 2, and from the first even term to the 1002nd there are 1001 steps of -2, a total drop of 2002 — landing exactly on 2002 - 2002 = 0. The value sits right where the pattern points, and 0 is one of the offered choices, so (C) is consistent.
💡Key takeaway

When a sequence zig-zags, sort the terms by even and odd position — each half often becomes a steady step-by-a-fixed-amount pattern you can jump straight to.

  • Generate the first several terms
  • Split odd and even positions
  • Extend the even pattern to position 2004
  • Match the value to a choice

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