AMC 10 · 2004 · #19
Grade 6 arithmeticIn the sequence 2001, 2002, 2003, … , 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+2002−2003=2000. What is the
2004th term in this sequence?
Pick an answer.
AMC 10 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A sequence starts $2001, 2002, 2003, \ldots$ From the fourth term on, each term equals the sum of the two terms before the one just ahead of it, minus that one just-ahead term: $a_n = a_{n-3} + a_{n-2} - a_{n-1}$. For instance $a_4 = 2001 + 2002 - 2003 = 2000$. Find the $2004\text{th}$ term.
Givens: The first three terms are $a_1 = 2001$, $a_2 = 2002$, $a_3 = 2003$.; The rule is $a_n = a_{n-3} + a_{n-2} - a_{n-1}$: add the two terms that come before the previous term, then subtract the previous term.; The worked example confirms the rule: $a_4 = 2001 + 2002 - 2003 = 2000$.; Answer choices: (A) $-2004$, (B) $-2$, (C) $0$, (D) $4003$, (E) $6007$
Unknowns: The value of the $2004\text{th}$ term, $a_{2004}$.
Understand
Restated: A sequence starts $2001, 2002, 2003, \ldots$ From the fourth term on, each term equals the sum of the two terms before the one just ahead of it, minus that one just-ahead term: $a_n = a_{n-3} + a_{n-2} - a_{n-1}$. For instance $a_4 = 2001 + 2002 - 2003 = 2000$. Find the $2004\text{th}$ term.
Givens: The first three terms are $a_1 = 2001$, $a_2 = 2002$, $a_3 = 2003$.; The rule is $a_n = a_{n-3} + a_{n-2} - a_{n-1}$: add the two terms that come before the previous term, then subtract the previous term.; The worked example confirms the rule: $a_4 = 2001 + 2002 - 2003 = 2000$.; Answer choices: (A) $-2004$, (B) $-2$, (C) $0$, (D) $4003$, (E) $6007$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #15 Organize Information in More Ways, #3 Eliminate Possibilities
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 $2001, 2002, 2003, 2000, 2005, 1998, 2007, 1996, \ldots$ 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).
Execute — Answer: C
4.NBT.B.4 Step 1 Generate the first several terms
- Apply the rule $a_n = a_{n-3} + a_{n-2} - a_{n-1}$ repeatedly.
- $a_4 = 2001 + 2002 - 2003 = 2000$.
- $a_5 = 2002 + 2003 - 2000 = 2005$.
- $a_6 = 2003 + 2000 - 2005 = 1998$.
- $a_7 = 2000 + 2005 - 1998 = 2007$.
- $a_8 = 2005 + 1998 - 2007 = 1996$.
- So far the sequence is $2001, 2002, 2003, 2000, 2005, 1998, 2007, 1996, \ldots$
💡 Compute a handful of terms first — the pattern rarely shows itself until you can see several terms lined up.
4.OA.C.5 Step 2 Split odd and even positions
- The full list zig-zags, but sorting terms by their position number tames it.
- The odd positions are $a_1=2001,\; a_3=2003,\; a_5=2005,\; a_7=2007$ — going up by $2$ each time.
- The even positions are $a_2=2002,\; a_4=2000,\; a_6=1998,\; a_8=1996$ — going down by $2$ each time.
- Because $2004$ is even, only the even-position sequence matters, and it drops by exactly $2$ from one even term to the next.
💡 A messy sequence often hides two tidy ones — separating by position turns the zig-zag into a straight line.
6.EE.B.6 Step 3 Extend the even pattern to position 2004
- Number the even-position terms: the $1$st is $a_2=2002$, the $2$nd is $a_4=2000$, the $3$rd is $a_6=1998$, and in general the $k$th even term is $E_k = 2002 - 2(k-1)$, since it starts at $2002$ and loses $2$ each step.
- Position $2004$ is the $2004 \div 2 = 1002$nd even term, so $k=1002$.
- Then $E_{1002} = 2002 - 2(1002-1) = 2002 - 2\cdot 1001 = 2002 - 2002 = 0$.
💡 Once a sequence changes by a fixed step, one formula jumps straight to any term without listing them all.
4.OA.C.5 Step 4 Match the value to a choice
- The $2004\text{th}$ term equals $0$, which is choice (C).
- The distractors correspond to unrelated miscounts: $4003$ (D) is a pair-sum $a_1+a_2$, $-2004$ (A) and $-2$ (B) come from mis-shifting the pattern into negatives, and $6007$ (E) has no basis here.
- Only $0$ matches the even-position rule, so the answer is (C).
💡 A single trusted value knocks out every other choice, so the exact term settles the answer.
4.NBT.B.4 Apply the rule $a_n = a_{n-3} + a_{n-2} - a_{n-1}$ repeatedly. $a_4 = 2001 + 200 4.OA.C.5 The full list zig-zags, but sorting terms by their position number tames it. The 6.EE.B.6 Number the even-position terms: the $1$st is $a_2=2002$, the $2$nd is $a_4=2000$ 4.OA.C.5 The $2004\text{th}$ term equals $0$, which is choice (C). The distractors corres Review
Reasonableness: The even terms $2002, 2000, 1998, \ldots$ march steadily downward by $2$, and from the first even term to the $1002$nd 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.
Alternative: Rewrite the rule as $a_n + a_{n-1} = a_{n-2} + a_{n-3}$: the sum of two neighboring terms repeats every two steps. For even $n$ this sum is always $a_2 + a_1 = 4003$, so $a_{2003} + a_{2004} = 4003$. The odd terms climb by $2$ from $2001$, giving $a_{2003} = 2001 + 2\cdot 1001 = 4003$. Therefore $a_{2004} = 4003 - 4003 = 0$, confirming (C).
CCSS standards used (min grade 6)
4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Computing the early terms $a_4$ through $a_8$ from the recurrence to expose the pattern.)4.OA.C.5Generate a number or shape pattern following a given rule (Separating odd and even positions and recognizing the even-position terms decrease by 2, then confirming the final value against the choices.)6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Writing the general even-term expression $E_k = 2002 - 2(k-1)$ and evaluating it at $k=1002$ to reach the 2004th term directly.)
⭐ 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.
⭐ 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.
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