Competition · AMC preparation · step 4 of 4
AMC 8 · 2013 · #3
Grade 4 arithmeticpatternPick an answer.
AMC 8 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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 (-, +, -, +, …), 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 (-odd) + (next even) — solve one of them, and reuse the result for all the others. Once S is known, the final 4 · S is a one-line multiplication.
Pair up the terms
Group the 1000 terms into consecutive pairs — each shaped (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.
Grouping the alternating sum -1+2-3+4-…+1000 into consecutive pairs collapses it to 500.
▸ Why?
You may add the 1000 terms in any grouping, so bracketing them as (-1+2)+(-3+4)+…+(-999+1000) leaves the total unchanged.
▸ Why?
When you add a string of numbers, pushing any two together first and the rest later always reaches the same total.
▸ Why?
Each bracket (a negative odd plus the next even) equals 1, because the even number is one more than the odd number just below it, so taking that odd amount away leaves exactly 1.
▸ Why?
Removing the odd number undoes the same odd amount tucked inside the even number, cancelling it to nothing.
▸ Why?
Once the odd part cancels to zero, only the leftover 1 stays, since adding it to zero keeps it as 1.
▸ Why?
There are exactly 500 brackets, because 1000 terms taken two at a time form 1000 ÷ 2 = 500 equal groups.
▸ Why?
Sorting a total into equal groups of a fixed size gives a group count equal to the total divided by that size — here 500 groups of 2 rebuild 1000.
▸ Why?
Adding those 500 ones gives 500, since 500 equal groups of 1 is 500 × 1.
Check what each pair gives
Compute a few pairs — every one collapses the same way: -(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.C.5Look For A PatternCount the pairs
Count the pairs: 1000 terms ÷ 2 per pair = 500 pairs, so 500 ones get added.
Knowing how many copies of the repeating chunk fit into the whole is the second half of pattern reasoning.
4.OA.A.3Look For A PatternAdd the pairs and multiply by 4
Sum the 500 ones to get S = 500, then multiply by the outside 4: 4 · 500 = 2000.
Once the pattern collapses the sum to 500, the rest is a one-step multiplication of multi-digit whole numbers — Grade 4 territory.
4.NBT.B.5Look For A PatternThis AMC 8 problem only needs Grade 4 pattern-finding — pair up the terms, notice each pair adds to 1, then multiply!
- Pair up the terms
- Check what each pair gives
- Count the pairs
- Add the pairs and multiply by 4
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