AMC 10 · 2003 · #1
Grade 3 arithmeticWhat is the difference between the sum of the first 2003 even counting numbers and the sum of the first 2003 odd counting numbers?
Pick an answer.
AMC 10 2003 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: Take the sum of the first $2003$ even counting numbers and subtract the sum of the first $2003$ odd counting numbers. Report the difference.
Givens: The first $2003$ even counting numbers: $2, 4, 6, \dots, 4006$; The first $2003$ odd counting numbers: $1, 3, 5, \dots, 4005$; Answer choices: (A) $0$, (B) $1$, (C) $2$, (D) $2003$, (E) $4006$
Unknowns: The value of (sum of the even numbers) $-$ (sum of the odd numbers)
Understand
Restated: Take the sum of the first $2003$ even counting numbers and subtract the sum of the first $2003$ odd counting numbers. Report the difference.
Givens: The first $2003$ even counting numbers: $2, 4, 6, \dots, 4006$; The first $2003$ odd counting numbers: $1, 3, 5, \dots, 4005$; Answer choices: (A) $0$, (B) $1$, (C) $2$, (D) $2003$, (E) $4006$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities
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.
Execute — Answer: D
3.OA.D.9 Step 1 Pair the two lists position by position
- Instead of adding each whole list first, line the lists up and subtract in matching positions.
- The first even number pairs with the first odd number, the second with the second, and so on.
- Because subtraction across the two full sums can be regrouped, (even sum) $-$ (odd sum) equals the sum of these matched differences: $(2-1)+(4-3)+(6-5)+\dots+(4006-4005)$.
💡 Every even number sits exactly one step past the odd number in the same position, so compare them one pair at a time.
2.OA.B.2 Step 2 Every pair differs by exactly 1
- Work out any single pair: $2-1=1$, $4-3=1$, $6-5=1$, all the way to $4006-4005=1$.
- Each even number is exactly $1$ more than the odd number lined up with it, so every one of the matched differences equals $1$.
- There are no exceptions and nothing changes along the way.
💡 Consecutive whole numbers are always $1$ apart, so each even-minus-odd pair is just $1$.
3.OA.A.1 Step 3 Add up the 2003 ones
- The whole difference is now $2003$ copies of $1$ added together.
- That is $2003 \times 1 = 2003$.
- So the sum of the first $2003$ even numbers is exactly $2003$ more than the sum of the first $2003$ odd numbers, which is choice (D).
💡 Adding one $2003$ times is the same as multiplying, and $2003\times 1$ is just $2003$.
3.OA.D.9 Instead of adding each whole list first, line the lists up and subtract in match 2.OA.B.2 Work out any single pair: $2-1=1$, $4-3=1$, $6-5=1$, all the way to $4006-4005=1 3.OA.A.1 The whole difference is now $2003$ copies of $1$ added together. That is $2003 \ Review
Reasonableness: The even list is bigger than the odd list, so the difference must be positive — that kills (A) $0$. Each matched pair differs by $1$, not $2$, so the total gap is one per pair, not two; that rules out (E) $4006$, which is what you would get if each pair differed by $2$. With $2003$ pairs each contributing $1$, the difference is $2003$, matching (D).
Alternative: Use sum formulas. The first $n$ even numbers sum to $n(n+1)$ and the first $n$ odd numbers sum to $n^2$. Their difference is $n(n+1)-n^2 = n$. With $n=2003$ the difference is $2003$, choice (D) again.
CCSS standards used (min grade 3)
3.OA.D.9Identify arithmetic patterns and explain using properties of operations (Regrouping the two big sums into matched pairs and seeing the even number is always one step past the odd number in the same position.)2.OA.B.2Fluently add and subtract within 20 using mental strategies (Computing each matched difference, e.g. $2-1$, $4-3$, and seeing every pair equals $1$.)3.OA.A.1Interpret products of whole numbers as total number of objects in groups (Reading $2003$ copies of $1$ as $2003\times 1 = 2003$ to get the final difference.)
⭐ Instead 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 $2003$.
⭐ Instead 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 $2003$.
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