AMC 10 · 2020 · #8
Easy mode Grade 4Problem
Picture a long string of additions and subtractions using the numbers 1,2,3,…,200 in order.
The pattern works in chunks of four numbers. In each chunk, the first three numbers get a plus sign and the fourth number gets a minus sign. So the expression starts 1+2+3−4, then continues +5+6+7−8, and keeps going the same way all the way to ⋯+197+198+199−200.
What is the total value of the whole expression?
1+2+3−4+5+6+7−8+⋯+197+198+199−200
Pick an answer.
AMC 10 2020 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 $1 + 2 + 3 - 4 + 5 + 6 + 7 - 8 + \cdots + 197 + 198 + 199 - 200$. Three terms are added, then one subtracted — and the cycle repeats through $200$.
Givens: Terms run through $1, 2, 3, \ldots, 200$ in order; Sign pattern: $+, +, +, -, +, +, +, -, \ldots$; Every $4$th term is subtracted (the multiples of $4$); Choices: (A) $9{,}800$, (B) $9{,}900$, (C) $10{,}000$, (D) $10{,}100$, (E) $10{,}200$
Unknowns: The value of the entire alternating sum
Understand
Restated: Compute $1 + 2 + 3 - 4 + 5 + 6 + 7 - 8 + \cdots + 197 + 198 + 199 - 200$. Three terms are added, then one subtracted — and the cycle repeats through $200$.
Givens: Terms run through $1, 2, 3, \ldots, 200$ in order; Sign pattern: $+, +, +, -, +, +, +, -, \ldots$; Every $4$th term is subtracted (the multiples of $4$); Choices: (A) $9{,}800$, (B) $9{,}900$, (C) $10{,}000$, (D) $10{,}100$, (E) $10{,}200$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #7 Identify Subproblems, #9 Solve an Easier Related Problem, #3 Eliminate Possibilities
The sign pattern repeats every $4$ terms — Tool #5 says: group terms by the repeating unit and compute each block. Tool #7 then breaks the full expression into the sum of $50$ block values, and finally the sum of an arithmetic sequence ($2, 10, 18, \ldots$). Tool #9 (easier) lets us verify the per-block formula on just $1, 2, 3$ blocks before sweeping all $50$. This combination avoids any algebraic variable.
Execute — Answer: B
4.OA.C.5 Step 1 - Group the $200$ terms into blocks of $4$: $(1+2+3-4), (5+6+7-8), (9+10+11-12), \ldots, (197+198+199-200)$.
- There are $200 / 4 = 50$ such blocks.
💡 The pattern repeats every $4$ — group along the repeat.
2.OA.A.1 Step 2 - Compute the first few blocks to spot the per-block pattern.
- Block $k$ contains the four terms $4k-3, 4k-2, 4k-1, 4k$ with the last one subtracted.
💡 Try $k = 1, 2, 3$ — small cases reveal the rule.
4.OA.C.5 Step 3 - The block totals $2, 10, 18, \ldots$ rise by $8$ each step (each successive block shifts all four terms up by $4$, so the net sum shifts by $4 \cdot (1+1+1-1) = 4 \cdot 2 = 8$).
- Last block ($k = 50$): $197+198+199-200 = 394$.
💡 Each block is $8$ more than the previous one — arithmetic progression.
4.NBT.B.5 Step 4 - Sum the arithmetic sequence $2, 10, 18, \ldots, 394$ ($50$ terms, first $= 2$, last $= 394$).
- Pair first with last: $2 + 394 = 396$, and there are $25$ such pairs.
💡 Pair the $50$ block values from both ends — every pair sums to the same $396$.
4.NBT.B.5 Step 5 Compute $25 \cdot 396$ by splitting $396 = 400 - 4$.
💡 $25 \cdot 400$ is easy ($10000$); subtract $100$ for the $-4$ piece.
4.NBT.A.2 Step 6 $9{,}900$ matches choice (B).
💡 Read the matching answer choice.
4.OA.C.5 Group the $200$ terms into blocks of $4$: $(1+2+3-4), (5+6+7-8), (9+10+11-12), \ 2.OA.A.1 Compute the first few blocks to spot the per-block pattern. Block $k$ contains t 4.OA.C.5 The block totals $2, 10, 18, \ldots$ rise by $8$ each step (each successive bloc 4.NBT.B.5 Sum the arithmetic sequence $2, 10, 18, \ldots, 394$ ($50$ terms, first $= 2$, l 4.NBT.B.5 Compute $25 \cdot 396$ by splitting $396 = 400 - 4$. 4.NBT.A.2 $9{,}900$ matches choice (B). Review
Reasonableness: Quick sanity check by another grouping. The full alternating expression equals $(1 + 2 + \cdots + 200) - 2 \cdot (4 + 8 + 12 + \cdots + 200)$ — because every multiple of $4$ appears with a minus sign instead of a plus, costing twice its value. Compute: $\sum_{1}^{200} = \dfrac{200 \cdot 201}{2} = 20100$. Multiples of $4$ up to $200$: there are $50$ of them, summing to $4 \cdot (1 + 2 + \cdots + 50) = 4 \cdot 1275 = 5100$. So the answer is $20100 - 2 \cdot 5100 = 20100 - 10200 = 9900$. ✓
Alternative: Tool #16 (Change Focus / Complement): instead of grouping in fours, start from the all-positive total $1 + 2 + \cdots + 200 = 20{,}100$ and subtract twice the multiples of $4$ (each $4k$ should have been $+4k$ but is $-4k$, a swing of $-2 \cdot 4k$). This is the verification path above and lands on the same $9{,}900$.
CCSS standards used (min grade 4)
4.OA.C.5Generate a number or shape pattern following a given rule (Recognizing the $+,+,+,-$ cycle of length $4$ and the resulting arithmetic block-total pattern $2, 10, 18, \ldots$)2.OA.A.1Solve one- and two-step word problems using addition and subtraction within 100 (Computing the first few block values ($1+2+3-4 = 2$, etc.).)4.NBT.B.5Multiply a whole number of up to four digits by a one-digit whole number (Pairing the $50$ block values into $25 \cdot 396$ and splitting $25 \cdot 396 = 25 \cdot 400 - 25 \cdot 4$ for easy mental arithmetic.)4.NBT.A.2Read and write multi-digit whole numbers and compare using symbols (Matching the computed value $9{,}900$ to choice (B).)
⭐ This AMC 10 problem only needs Grade 4 pattern-spotting you already know — group every $4$ terms: $(1+2+3-4) = 2$, $(5+6+7-8) = 10$, $(9+10+11-12) = 18$, jumping by $8$ each time. Add up the $50$ block totals (last one is $394$) to get $9{,}900$.
⭐ This AMC 10 problem only needs Grade 4 pattern-spotting you already know — group every $4$ terms: $(1+2+3-4) = 2$, $(5+6+7-8) = 10$, $(9+10+11-12) = 18$, jumping by $8$ each time. Add up the $50$ block totals (last one is $394$) to get $9{,}900$.
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