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AMC 10 · 2020 · #8

Easy mode Grade 4
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Problem

Picture a long string of additions and subtractions using the numbers 1,2,3,…,2001, 2, 3, \ldots, 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+3−41+2+3-4, then continues +5+6+7−8+5+6+7-8, and keeps going the same way all the way to ⋯+197+198+199−200\cdots +197+198+199-200.

What is the total value of the whole expression?

1+2+3−4+5+6+7−8+⋯+197+198+199−2001+2+3-4+5+6+7-8+\cdots+197+198+199-200

(A) 9,800(B) 9,900(C) 10,000(D) 10,100(E) 10,200\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 $200$.

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

#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 $9{,}900$.

⭐ 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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