Competition · AMC preparation · step 4 of 4
AMC 8 · 2018 · #5
Grade 4 arithmeticpatternPick an answer.
AMC 8 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The expression is huge, but it has a very regular structure: alternating +odd, -even. Tool #5 (Look for a Pattern) is the natural fit — re-group the terms as (1-2)+(3-4)+(5-6)+… and notice that every pair has the same value. Tool #9 (Solve an Easier Related Problem) confirms the pattern with a tiny version (e.g., 1+3+5 - 2 - 4) before trusting it on the big one. Tool #7 (Identify Subproblems) helps us split the work cleanly into: (a) the leftover term 2019, (b) the paired part, and (c) counting how many pairs there are.
Try a smaller sum
Test it small first: 1+3+5-2-4 regroups as (1-2)+(3-4)+5 = 3 — each pair matches and the last odd number is left with no partner.
Trying a tiny version first is a Grade 4 "generate and analyze a pattern" move — it shows the structure before we commit.
4.OA.C.5Solve An Easier Related ProblemPair the terms
Regroup the whole expression the same way: (1-2)+(3-4)+…+(2017-2018) with 2019 left unpaired at the very end.
Splitting the expression into "paired part" plus a single leftover term is a Grade 4 multi-step word-problem move.
The whole expression equals the paired blocks (1-2)+(3-4)+(5-6)+…+(2017-2018) with 2019 left over, and rearranging the terms into this shape does not change the value.
▸ Why?
Interleaving each odd number with the even number right after it just reorders the same adds and take-aways, and doing them in a different order lands on the same running total.
▸ Why?
Once the terms are lined up as odd-then-even neighbors, we are free to combine each neighbor pair first before adding the blocks together, and which terms you push together first never changes the total.
▸ Why?
The final odd number 2019 has no even partner, so it stays as its own leftover block, and the paired blocks together with that leftover still cover every original term exactly once, with nothing added or dropped.
Find each pair's value
Each pair is (odd) - (next even) = odd - (odd+1) = -1, so every pair falls short by 1.
Spotting that every pair gives the same shortfall is exactly the Grade 4 "find the rule of the pattern" idea.
4.OA.C.5Look For A PatternCount the pairs
Count the pairs: each uses one even number from {2,4,…,2018}, and dividing by 2 gives {1,…,1009}, so there are 1009 pairs.
Counting how many even numbers fit up to 2018 is a Grade 4 multiples / factor-pair skill.
4.OA.B.4Identify SubproblemsSubtract from 2019
Finish in one line: subtract the 1009 pair-shortfalls from the leftover 2019, i.e. 2019 - 1009, which gives the answer.
Subtracting a 4-digit number from a 4-digit number to finish is Grade 4 multi-digit subtraction.
4.NBT.B.4Look For A PatternThis AMC 8 problem only needs Grade 4 pattern-finding and multi-digit subtraction you already know!
- Try a smaller sum
- Pair the terms
- Find each pair's value
- Count the pairs
- Subtract from 2019
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