AMC 10 · 2004 · #12
Grade 6 arithmeticPick an answer.
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).
Generate the first several terms
Generating terms gives 2000, 2005, 1998, 2007, 1996 — a zig-zag.
Compute a handful of terms first — the pattern rarely shows itself until you can see several terms lined up.
4.NBT.B.4Look For A PatternSplit odd and even positions
Split by position: odds climb by 2, evens drop by 2.
A messy sequence often hides two tidy ones — separating by position turns the zig-zag into a straight line.
Separating the odd positions from the even ones turns one zig-zag sequence into two straight ones.
▸ Why?
The rule treats every other term the same way, so the positions fall into two alternating classes.
▸ Why?
Inside each class the terms change by the same fixed amount, so one formula jumps straight to any position.
Extend the even pattern to position 2004
Position 2004 is the 1002nd even term, which lands on 0.
Once a sequence changes by a fixed step, one formula jumps straight to any term without listing them all.
6.EE.B.6Look For A PatternMatch the value to a choice
No other choice fits the even-position rule, so the answer is 0, choice (C).
A single trusted value knocks out every other choice, so the exact term settles the answer.
4.OA.C.5Eliminate PossibilitiesWhen 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