AMC 10 · 2020 · #8

Grade 4 arithmetic

Archetype Arithmetic Expression Decomposition

pattern-recognitionsequences-arithmeticmental-arithmetic pattern-recognitionidentify-subproblems ↑ Prerequisites: sequences-arithmetic

📏 Medium solution 💡 3 insights

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Problem

What is the value of $1+2+3-4+5+6+7-8+\cdots+197+198+199-200?$

$\textbf{(A) } 9,800 \qquad \textbf{(B) } 9,900 \qquad \textbf{(C) } 10,000 \qquad \textbf{(D) } 10,100 \qquad \textbf{(E) } 10,200$

Pick an answer.

(A)

9,800

(B)

9,900

(C)

10,000

(D)

10,100

(E)

10,200

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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 $200$.

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

#5 Look for a Pattern 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.

$$\underbrace{(1+2+3-4)}_{\text{block 1}} + \underbrace{(5+6+7-8)}_{\text{block 2}} + \cdots + \underbrace{(197+198+199-200)}_{\text{block 50}}$$

💡 The pattern repeats every $4$ — group along the repeat.

#9 Solve an Easier Related Problem 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.

$$1+2+3-4 = 2,\quad 5+6+7-8 = 10,\quad 9+10+11-12 = 18$$

💡 Try $k = 1, 2, 3$ — small cases reveal the rule.

#5 Look for a Pattern 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 ($k = 50$): $197+198+199-200 = 394$.

$$\text{Block }k\text{ value} = 8k - 6,\quad\text{Block }50 = 8 \cdot 50 - 6 = 394$$

💡 Each block is $8$ more than the previous one — arithmetic progression.

#7 Identify Subproblems 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.

$$S = \dfrac{50}{2} \cdot (2 + 394) = 25 \cdot 396$$

💡 Pair the $50$ block values from both ends — every pair sums to the same $396$.

#7 Identify Subproblems 4.NBT.B.5 Step 5

Compute $25 \cdot 396$ by splitting $396 = 400 - 4$.

$$25 \cdot 396 = 25 \cdot 400 - 25 \cdot 4 = 10000 - 100 = 9900$$

💡 $25 \cdot 400$ is easy ($10000$); subtract $100$ for the $-4$ piece.

#3 Eliminate Possibilities 4.NBT.A.2 Step 6

$9{,}900$ matches choice (B).

$$9{,}900\;\Rightarrow\;\textbf{(B)}$$

💡 Read the matching answer choice.

[1] #5 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] #9 2.OA.A.1 Compute the first few blocks to spot the per-block pattern. Block $k$ contains t

[3] #5 4.OA.C.5 The block totals $2, 10, 18, \ldots$ rise by $8$ each step (each successive bloc

[4] #7 4.NBT.B.5 Sum the arithmetic sequence $2, 10, 18, \ldots, 394$ ($50$ terms, first $= 2$, l

[5] #7 4.NBT.B.5 Compute $25 \cdot 396$ by splitting $396 = 400 - 4$.

[6] #3 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.5 Generate 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.1 Solve 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.5 Multiply 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.2 Read 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 $9{,}900$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 4)

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