AMC 8 · 2013 · #3

Easy mode Grade 4

Problem

Imagine the numbers from $1$ all the way up to $1000$ , written in a long row.

We are going to add and subtract them in a pattern. The odd numbers ( $1, 3, 5, \ldots, 999$ ) each get a minus sign. The even numbers ( $2, 4, 6, \ldots, 1000$ ) each get a plus sign.

So the expression inside the parentheses looks like this:

$-1+2-3+4-5+6-7+\cdots+1000$

Then we take that total and multiply it by $4$ .

What is the value of $4 \cdot (-1+2-3+4-5+6-7+\cdots+1000)$ ?

Pick an answer.

(A)

-10

(B)

(C)

(D)

500

(E)

2000

AMC 8 2013 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.

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Toolkit + CCSS Solution

Understand

Restated: Find the value of $4 \cdot (-1 + 2 - 3 + 4 - 5 + 6 - 7 + \cdots + 1000)$, where the signs inside the parentheses alternate $-, +, -, +, \ldots$ from $1$ all the way up to $1000$.

Givens: An alternating sum $S = -1 + 2 - 3 + 4 - 5 + 6 - \cdots - 999 + 1000$; Every odd number ($1, 3, 5, \ldots, 999$) is subtracted; every even number ($2, 4, 6, \ldots, 1000$) is added; The whole sum $S$ is multiplied by $4$; Answer choices: (A) $-10$, (B) $0$, (C) $1$, (D) $500$, (E) $2000$

Unknowns: The numerical value of $4 \cdot S$

Understand

Restated: Find the value of $4 \cdot (-1 + 2 - 3 + 4 - 5 + 6 - 7 + \cdots + 1000)$, where the signs inside the parentheses alternate $-, +, -, +, \ldots$ from $1$ all the way up to $1000$.

Plan

Primary tool: #5 Look for a Pattern

Secondary: #7 Identify Subproblems

Adding $1000$ signed numbers one by one is hopeless — we need structure. Tool #5 (Look for a Pattern) is the perfect fit because the signs repeat in a clear two-step cycle ($-, +, -, +, \ldots$), which means consecutive pairs of terms might collapse to something simple. Tool #7 (Identify Subproblems) makes that idea concrete: we split $S$ into $500$ small subproblems — each one is a pair $(-\text{odd}) + (\text{next even})$ — solve one of them, and reuse the result for all the others. Once $S$ is known, the final $4 \cdot S$ is a one-line multiplication.

Execute — Answer: E

#7 Identify Subproblems 4.OA.A.3 Step 1

Pair up the terms

Group the $1000$ terms inside the parentheses into consecutive pairs.
The signs alternate $-, +$, so each pair has the form (negative odd) + (next even).

$$S = (-1 + 2) + (-3 + 4) + (-5 + 6) + \cdots + (-999 + 1000)$$

💡 Splitting one giant sum into many small two-number sums is the Tool #7 subproblems move, and it works because the sign pattern lines up perfectly with pairs.

#5 Look for a Pattern 4.OA.C.5 Step 2

Check what each pair gives

Evaluate the first few pairs to see what each subproblem evaluates to.
Every pair looks like $(-(2k-1)) + 2k = 1$.

$$-1 + 2 = 1, \quad -3 + 4 = 1, \quad -5 + 6 = 1, \quad \ldots, \quad -999 + 1000 = 1$$

💡 Computing a few terms and seeing the same answer pop out is exactly the Grade 4 "generate and analyze patterns" habit.

#5 Look for a Pattern 4.OA.A.3 Step 3

Count the pairs

Count the pairs.
The parentheses contain $1000$ terms, and each pair uses $2$ terms, so the number of pairs (and therefore the number of $1$'s being added) is $1000 \div 2 = 500$.

$$\text{number of pairs} = \dfrac{1000}{2} = 500$$

💡 Knowing how many copies of the repeating chunk fit into the whole is the second half of pattern reasoning.

#5 Look for a Pattern 4.NBT.B.5 Step 4

Add the pairs and multiply by 4

Add up the $500$ copies of $1$ to get $S$, then multiply by the $4$ that sits outside the parentheses.

$$S = 500 \times 1 = 500, \quad 4 \cdot S = 4 \times 500 = 2000 \;\Rightarrow\; \textbf{(E)}$$

💡 Once the pattern collapses the sum to $500$, the rest is a one-step multiplication of multi-digit whole numbers — Grade 4 territory.

[1] #7 4.OA.A.3 Group the $1000$ terms inside the parentheses into consecutive pairs. The signs

[2] #5 4.OA.C.5 Evaluate the first few pairs to see what each subproblem evaluates to. Every pai

[3] #5 4.OA.A.3 Count the pairs. The parentheses contain $1000$ terms, and each pair uses $2$ te

[4] #5 4.NBT.B.5 Add up the $500$ copies of $1$ to get $S$, then multiply by the $4$ that sits ou

Review

Reasonableness: A sanity check: the $1000$ terms split into $500$ pairs that each sum to $+1$, so $S = 500$. Multiplying by $4$ gives $2000$, which matches choice (E). The answer is positive (good — every pair was positive), and its size ($2000$) sits at the top of the answer list, which makes sense because we are scaling a $500$-pair sum by $4$.

Alternative: Tool #16 (Change Focus): instead of adding term-by-term, separate the series into the evens and the odds. The evens $2 + 4 + 6 + \cdots + 1000 = 2(1 + 2 + \cdots + 500) = 2 \cdot \tfrac{500 \cdot 501}{2} = 250{,}500$. The odds $1 + 3 + 5 + \cdots + 999 = 500^2 = 250{,}000$. Then $S = (\text{evens}) - (\text{odds}) = 250{,}500 - 250{,}000 = 500$, and $4 \cdot 500 = 2000$ — same answer (E).

CCSS standards used (min grade 4)

4.OA.A.3 Solve multi-step word problems with whole numbers using the four operations (Splitting the $1000$-term expression into $500$ pair-subproblems and then counting those pairs ($1000 \div 2 = 500$) is exactly the multi-step whole-number reasoning of this standard.)
4.OA.C.5 Generate and analyze patterns (Recognizing that every consecutive pair $(-1+2), (-3+4), (-5+6), \ldots$ evaluates to the same value $1$ — i.e., spotting and using a repeating numerical pattern.)
4.NBT.B.5 Multiply a whole number of up to four digits by a one-digit whole number (Computing $500 \times 1 = 500$ (sum of $500$ copies of $1$) and $4 \times 500 = 2000$ (the final outside multiplication).)

⭐ This AMC 8 problem only needs Grade 4 pattern-finding — pair up the terms, notice each pair adds to $1$, then multiply!

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Pair up the terms

Check what each pair gives

Count the pairs

Add the pairs and multiply by 4

Review

CCSS standards used (min grade 4)

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