AMC 8 · 2015 · #9

Grade 3 arithmeticpattern
sequences-arithmeticpattern-recognitionperfect-squares pattern-recognitionidentify-subproblems ↑ Prerequisites: multi-digit-arithmeticsequences-arithmetic
📏 Medium solution 💡 3 insights
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Problem
Janabel sells 1 widget on day 1, 3 widgets on day 2, 5 widgets on day 3, and each day she sells 2 more than the day before. After working 20 days, how many widgets has she sold in total?

Pick an answer.

(A)
39
(B)
40
(C)
210
(D)
400
(E)
401

AMC 8 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Look for a Pattern

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.

1STEP 1

Start easy — add up the first few days to get the running totals 1, 4, 9, 16.

S₁ = 1, S₂ = 1+3 = 4, S₃ = 1+3+5 = 9, S₄ = 1+3+5+7 = 16
2STEP 2

Those totals are the perfect squares — the running total after n days equals .

S_n = n² for n = 1, 2, 3, 4
3STEP 3

Test one more case — day 5 adds 9 widgets, giving 25 = 5², so the pattern holds.

S₅ = 16 + 9 = 25 = 5² ✓
4STEP 4

Apply the pattern to n = 20 — the total after 20 days is 20².

S₂0 = 20² = 400
5STEP 5

Match to the choices — 400 is option (D), and no other choice is 20 squared.

400 → (D)
Answer
400
Day 20 alone has 1 + 19 × 2 = 39 widgets, the average day sells 1+392\frac{1 + 39}{2} = 20 widgets, and 20 days × 20 widgets = 400. Same answer from a completely different route, so 400 is right. Choice (A) 39 is the day-20 number alone, and (E) 401 is an off-by-one trap; both are plausible distractors but neither is the total.
💡Key takeaway

This 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!