2004 AMC 10B Problem 19

The rule is $a_{n+3} = a_n + a_{n+1} - a_{n+2}$. Move the last term across:
$$a_{n+2} + a_{n+3} = a_n + a_{n+1}.$$
So the sum of two neighbors is unchanged when you slide the pair two places to the right. Starting from an odd index, every pair sums to $a_1 + a_2 = 4003$; starting from an even index, every pair sums to $a_2 + a_3 = 4005$.

For any even index $2k$, use both facts:
$$a_{2k} + a_{2k+1} = 4005, \qquad a_{2k+1} + a_{2k+2} = 4003.$$
Subtracting, $a_{2k+2} = a_{2k} - 2$. The even-indexed terms form an arithmetic progression with difference $-2$: $a_2 = 2002$, $a_4 = 2000$, $a_6 = 1998$, and in general $a_{2k} = 2002 - 2(k - 1) = 2004 - 2k$.

With $2k = 2004$, $k = 1002$ and $a_{2004} = 2004 - 2004 = 0$.

The answer is $\boxed{\textbf{(C)}\ 0}$.

A recurrence with coefficients $+1, +1, -1$ hides a conserved quantity; rearranging it to put a matching expression on both sides reveals the invariant "sum of adjacent terms, shifted by two." Once an invariant is found, distant terms are reachable by simple arithmetic instead of by iterating. Computing the first several terms and watching every other one is a good way to guess the same structure.

Just compute: $2001, 2002, 2003, 2000, 2005, 1998, 2007, 1996, \dots$. The even positions read $2002, 2000, 1998, 1996$, decreasing by $2$, while the odd positions climb by $2$. Position $2004$ is the $1002$nd even position, so its value is $2002 - 2 \cdot 1001 = 0$.

• An off-by-one in the number of -2 steps from a_2 to a_{2004} (1001 steps, not 1002), which produces -2, choice (B).
• Reading the odd-position pattern ($+2$ each step) and applying it to an even position, which gives a value near $4003$.