AMC 8 · 2018 · #5
Grade 4 arithmeticpatternProblem
What is the value of 1+3+5+⋯+2017+2019−2−4−6−⋯−2016−2018?
Pick an answer.
AMC 8 2018 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: Compute the value of the expression $1+3+5+\cdots+2017+2019 \;-\; (2+4+6+\cdots+2016+2018)$ — that is, add all odd numbers from $1$ to $2019$, then subtract all even numbers from $2$ to $2018$.
Givens: Positive part: all odd numbers from $1$ up to $2019$; Negative part: all even numbers from $2$ up to $2018$; Both lists are arithmetic sequences with common difference $2$; Answer choices: (A) $-1010$, (B) $-1009$, (C) $1008$, (D) $1009$, (E) $1010$
Unknowns: The exact value of the combined expression
Understand
Restated: Compute the value of the expression $1+3+5+\cdots+2017+2019 \;-\; (2+4+6+\cdots+2016+2018)$ — that is, add all odd numbers from $1$ to $2019$, then subtract all even numbers from $2$ to $2018$.
Givens: Positive part: all odd numbers from $1$ up to $2019$; Negative part: all even numbers from $2$ up to $2018$; Both lists are arithmetic sequences with common difference $2$; Answer choices: (A) $-1010$, (B) $-1009$, (C) $1008$, (D) $1009$, (E) $1010$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #9 Solve an Easier Related Problem, #7 Identify Subproblems
The expression is huge, but it has a very regular structure: alternating $+$odd, $-$even. Tool #5 (Look for a Pattern) is the natural fit — re-group the terms as $(1-2)+(3-4)+(5-6)+\cdots$ and notice that every pair has the same value. Tool #9 (Solve an Easier Related Problem) confirms the pattern with a tiny version (e.g., $1+3+5 - 2 - 4$) before trusting it on the big one. Tool #7 (Identify Subproblems) helps us split the work cleanly into: (a) the leftover term $2019$, (b) the paired part, and (c) counting how many pairs there are.
Execute — Answer: E
4.OA.C.5 Step 1 - Try the same idea on a small version first.
- Take $1+3+5 - 2 - 4$.
- Re-group as $(1-2)+(3-4)+5 = (-1)+(-1)+5 = 3$.
- Each odd-then-even pair contributes the same small amount, and the final odd number is left over without a partner.
- We will reuse this structure for the full problem.
💡 Trying a tiny version first is a Grade 4 "generate and analyze a pattern" move — it shows the structure before we commit.
4.OA.A.3 Step 2 - Re-group the full expression by pairing each positive odd number with the next negative even number, leaving $2019$ unpaired at the very end.
- This is the same shape we just verified on the small case.
💡 Splitting the expression into "paired part" plus a single leftover term is a Grade 4 multi-step word-problem move.
4.OA.C.5 Step 3 - Look at the value of one pair.
- Every pair has the form $(\text{odd}) - (\text{next even}) = \text{odd} - (\text{odd}+1)$, which is short by $1$.
- So each pair removes $1$ from the running total.
💡 Spotting that every pair gives the same shortfall is exactly the Grade 4 "find the rule of the pattern" idea.
4.OA.B.4 Step 4 - Count how many pairs there are.
- Each pair uses up one even number from $\{2,4,6,\ldots,2018\}$, so the number of pairs equals the number of even numbers in that list.
- Dividing every even number by $2$ gives $\{1,2,3,\ldots,1009\}$, so there are $1009$ pairs.
💡 Counting how many even numbers fit up to $2018$ is a Grade 4 multiples / factor-pair skill.
4.NBT.B.4 Step 5 - Combine everything.
- Instead of writing $1009$ copies of $-1$ and adding $2019$, think of it as: the leftover $2019$ has $1009$ pair-shortfalls of $1$ taken away from it.
- So the answer is $2019 - 1009$.
💡 Subtracting a 4-digit number from a 4-digit number to finish is Grade 4 multi-digit subtraction.
4.OA.C.5 Try the same idea on a small version first. Take $1+3+5 - 2 - 4$. Re-group as $( 4.OA.A.3 Re-group the full expression by pairing each positive odd number with the next n 4.OA.C.5 Look at the value of one pair. Every pair has the form $(\text{odd}) - (\text{ne 4.OA.B.4 Count how many pairs there are. Each pair uses up one even number from ${2,4,6, 4.NBT.B.4 Combine everything. Instead of writing $1009$ copies of $-1$ and adding $2019$, Review
Reasonableness: The odd sum has one extra term ($2019$) compared to the even sum, so we expect the answer to be near $2019$ minus a correction equal to the number of paired "shortfalls." Matching pairs $(1,2),(3,4),\ldots,(2017,2018)$ each cost $1$, and there are $1009$ of them, leaving $2019-1009=1010$. That is positive (because the extra odd term dominates) and just slightly above $1009$, which matches choice (E) and rules out the negative choices (A), (B).
Alternative: Tool #6 (Guess and Check) using sums of arithmetic series: sum of odd numbers $1+3+\cdots+2019 = 1010^2 = 1{,}020{,}100$ (there are $1010$ odd numbers and their sum is $n^2$). Sum of even numbers $2+4+\cdots+2018 = 2(1+2+\cdots+1009) = 2 \cdot \tfrac{1009 \cdot 1010}{2} = 1009 \cdot 1010 = 1{,}019{,}090$. The difference is $1{,}020{,}100 - 1{,}019{,}090 = 1010$, confirming (E).
CCSS standards used (min grade 4)
4.OA.C.5Generate a number or shape pattern following a given rule (Testing the pairing idea on the small case $1+3+5-2-4$ and recognizing that every $(\text{odd}-\text{even})$ pair contributes the same shortfall of $1$.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Splitting the long expression into "paired part" plus the leftover term $2019$, then combining the two pieces.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Counting the even numbers from $2$ to $2018$ as multiples of $2$, giving $2018 \div 2 = 1009$ pairs.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Finishing the calculation with the multi-digit subtraction $2019 - 1009 = 1010$.)
⭐ This AMC 8 problem only needs Grade 4 pattern-finding and multi-digit subtraction you already know!
⭐ This AMC 8 problem only needs Grade 4 pattern-finding and multi-digit subtraction you already know!
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