Competition · AMC preparation · step 4 of 4
AMC 8 · 2015 · #9
Grade 3 arithmeticpatternPick an answer.
AMC 8 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The daily sales 1, 3, 5, 7, … are the odd numbers, so this is really a question about the sum of the first 20 odd numbers. Tool #9 (Easier Problem) computes the running totals for small day counts (n = 1, 2, 3, 4). Tool #5 (Look for a Pattern) then spots that each running total is a perfect square (1, 4, 9, 16 = n²), so the answer for n = 20 is 20² — no long addition needed. Tool #3 (Eliminate) is a quick sanity check against the five answer choices.
Add up the first few days
Start easy — add up the first few days to get the running totals 1, 4, 9, 16.
Adding the first few odd numbers is just Grade 2 addition within 20 — small, safe, and it gives us real data to look at.
2.OA.B.2Solve An Easier Related ProblemSpot the square-number pattern
Those totals are the perfect squares — the running total after n days equals n².
Spotting a numerical pattern in the running totals is the Grade 3 "identify arithmetic patterns" standard in action.
The running total of widgets after n days is the sum of the first n odd numbers 1 + 3 + 5 + …, and that running total is always the perfect square n².
▸ Why?
Picture each widget as a dot and keep the dots arranged as a square. Start with one dot, a 1 × 1 square. If every day's dots exactly complete the next larger square, then after n days the dots form a full n × n square, whose count is n².
▸ Why?
Growing a square from (n-1) across to n across means wrapping an L-shaped border around it: a new row of (n-1) dots, a new column of (n-1) dots, and the single corner dot. Those three non-overlapping pieces make up the whole border, so it holds (n-1) + (n-1) + 1 = 2n - 1 dots.
▸ Why?
Janabel's day-n sale starts at 1 and adds 2 on each of the (n-1) following days, which is 1 + 2 × (n-1) = 2n - 1 — the very same count as that L-border, so each day's widgets are exactly the border that finishes the next square.
▸ Why?
A completed n × n square is n equal rows of n dots each, and n equal groups of n is n × n = n², so once day n finishes the square the running total is n².
Test the pattern once more
Test one more case — day 5 adds 9 widgets, giving 25 = 5², so the pattern holds.
A pattern from 4 cases can still fool you — testing one more case is the cheap insurance the toolkit asks for.
3.OA.D.9Look For A PatternApply the pattern to day 20
Apply the pattern to n = 20 — the total after 20 days is 20².
Squaring a multiple of 10 is a Grade 3 single-digit multiplication fact (2 × 2 = 4) with two extra zeros tacked on.
3.OA.C.7Look For A PatternMatch against the choices
Match to the choices — 400 is option (D), and no other choice is 20 squared.
Eliminating answer choices is the standard AMC multiple-choice safety check.
3.OA.C.7Eliminate PossibilitiesThis AMC 8 problem only needs the Grade 3 fact that the sum of the first n odd numbers is n² — which you can discover yourself with Grade 2 addition!
- Add up the first few days
- Spot the square-number pattern
- Test the pattern once more
- Apply the pattern to day 20
- Match against the choices
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