AMC 10 · 2003 · #1
Grade 3 arithmeticPick an answer.
AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Adding two lists of 2003 numbers and subtracting is huge and error-prone. Tool #5 (Look for a Pattern) spots that the two lists run in lockstep: 2 sits right after 1, 4 right after 3, and so on. So instead of two giant sums, pair the lists position by position and look at one pair at a time. Tool #7 (Identify Subproblems) turns the whole thing into 2003 tiny subtractions, each of which is the same. Tool #3 (Eliminate Possibilities) then checks the size of the answer against the choices.
Pair the two lists position by position
Line the lists up and subtract position by position: (even sum) - (odd sum) = (2-1)+(4-3)+…+(4006-4005).
Every even number sits exactly one step past the odd number in the same position, so compare them one pair at a time.
3.OA.D.9Look For A PatternEvery pair differs by exactly 1
Every matched pair gives the same gap of 1: 2-1=1, 4-3=1, …, 4006-4005=1.
Consecutive whole numbers are always 1 apart, so each even-minus-odd pair is just 1.
Consecutive whole numbers are always one apart, so each matched pair differs by exactly one.
▸ Why?
Each list climbs by the same fixed step, so partners in the same position stay a fixed distance apart.
▸ Why?
Pairing the two lists position by position matches every term with exactly one partner.
Add up the 2003 ones
The difference is 2003 copies of 1, so 2003 × 1 = 2003, which is choice (D).
Adding one 2003 times is the same as multiplying, and 2003 × 1 is just 2003.
3.OA.A.1Identify SubproblemsInstead of adding two long lists and subtracting, pair them up: each even number is just 1 past its odd partner, so 2003 pairs give a difference of 2003.
- Pair the two lists position by position
- Every pair differs by exactly 1
- Add up the 2003 ones