AMC 8 · 2013 · #3
Easy mode Grade 4Problem
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,…,999) each get a minus sign. The even numbers (2,4,6,…,1000) each get a plus sign.
So the expression inside the parentheses looks like this:
−1+2−3+4−5+6−7+⋯+1000
Then we take that total and multiply it by 4.
What is the value of 4⋅(−1+2−3+4−5+6−7+⋯+1000)?
Pick an answer.
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.
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 $1000$.
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) $2000$
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
4.OA.A.3 Step 1 - Group the $1000$ terms inside the parentheses into consecutive pairs.
- The signs alternate $-, +$, so each pair has the form (negative odd) + (next even).
💡 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.
4.OA.C.5 Step 2 - Evaluate the first few pairs to see what each subproblem evaluates to.
- Every pair looks like $(-(2k-1)) + 2k = 1$.
💡 Computing a few terms and seeing the same answer pop out is exactly the Grade 4 "generate and analyze patterns" habit.
4.OA.A.3 Step 3 - 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 $1000 \div 2 = 500$.
💡 Knowing how many copies of the repeating chunk fit into the whole is the second half of pattern reasoning.
4.NBT.B.5 Step 4 Add up the $500$ copies of $1$ to get $S$, then multiply by the $4$ that sits outside the parentheses.
💡 Once the pattern collapses the sum to $500$, the rest is a one-step multiplication of multi-digit whole numbers — Grade 4 territory.
4.OA.A.3 Group the $1000$ terms inside the parentheses into consecutive pairs. The signs 4.OA.C.5 Evaluate the first few pairs to see what each subproblem evaluates to. Every pai 4.OA.A.3 Count the pairs. The parentheses contain $1000$ terms, and each pair uses $2$ te 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.3Solve 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.5Generate 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.5Multiply 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!
⭐ This AMC 8 problem only needs Grade 4 pattern-finding — pair up the terms, notice each pair adds to $1$, then multiply!
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