AMC 8 · 2015 · #9
Easy mode Grade 3Problem
Janabel starts a new job selling widgets.
- On day 1, she sells 1 widget.
- On day 2, she sells 3 widgets.
- On day 3, she sells 5 widgets.
Each day she sells 2 more widgets than she sold the day before. So day 4 would be 7, day 5 would be 9, and so on.
How many widgets has Janabel sold in total after 20 days of work?
Pick an answer.
AMC 8 2015 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: 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?
Givens: Day $1$ sales $= 1$; Day $2$ sales $= 3$; Day $3$ sales $= 5$; Each next day, sales increase by $2$; Answer choices: (A) $39$, (B) $40$, (C) $210$, (D) $400$, (E) $401$
Unknowns: The total number of widgets sold over the first $20$ days
Understand
Restated: 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?
Givens: Day $1$ sales $= 1$; Day $2$ sales $= 3$; Day $3$ sales $= 5$; Each next day, sales increase by $2$; Answer choices: (A) $39$, (B) $40$, (C) $210$, (D) $400$, (E) $401$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #9 Solve an Easier Related Problem, #3 Eliminate Possibilities
The daily sales $1, 3, 5, 7, \dots$ 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 ($1, 4, 9, 16 = n^2$), so the answer for $n = 20$ is $20^2$ — no long addition needed. Tool #3 (Eliminate) is a quick sanity check against the five answer choices.
Execute — Answer: D
2.OA.B.2 Step 1 - Compute the running total for the first few days.
- This is the easier version of the same question ($n = 1, 2, 3, 4$ instead of $n = 20$).
💡 Adding the first few odd numbers is just Grade 2 addition within $20$ — small, safe, and it gives us real data to look at.
3.OA.D.9 Step 2 - Compare the running totals to a familiar pattern.
- The list $1, 4, 9, 16$ is exactly the perfect squares $1^2, 2^2, 3^2, 4^2$ — the running total after $n$ days equals $n^2$.
💡 Spotting a numerical pattern in the running totals is the Grade 3 "identify arithmetic patterns" standard in action.
3.OA.D.9 Step 3 - Check the pattern on one more case so we trust it before jumping to $n = 20$.
- Day $5$ adds $9$ widgets, so $S_5 = 16 + 9 = 25 = 5^2$.
- 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.C.7 Step 4 - Apply the pattern to $n = 20$.
- The total widgets sold after $20$ days is $20^2$.
💡 Squaring a multiple of $10$ is a Grade 3 single-digit multiplication fact ($2 \times 2 = 4$) with two extra zeros tacked on.
3.OA.C.7 Step 5 - Match to the answer choices.
- $400$ appears as option (D); none of the other choices are square numbers of $20$.
💡 Eliminating answer choices is the standard AMC multiple-choice safety check.
2.OA.B.2 Compute the running total for the first few days. This is the easier version of 3.OA.D.9 Compare the running totals to a familiar pattern. The list $1, 4, 9, 16$ is exac 3.OA.D.9 Check the pattern on one more case so we trust it before jumping to $n = 20$. Da 3.OA.C.7 Apply the pattern to $n = 20$. The total widgets sold after $20$ days is $20^2$. 3.OA.C.7 Match to the answer choices. $400$ appears as option (D); none of the other choi Review
Reasonableness: Day $20$ alone has $1 + 19 \times 2 = 39$ widgets, the average day sells $(1 + 39)/2 = 20$ widgets, and $20 \text{ days} \times 20 \text{ widgets} = 400$. 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.
Alternative: Tool #14 (Finite Differences) also works. The running totals $1, 4, 9, 16, 25, \dots$ have first differences $3, 5, 7, 9, \dots$ and constant second differences of $2$, signalling a quadratic formula $S_n = n^2$. Same answer, more machinery — the pattern route is faster for AMC 8 timing.
CCSS standards used (min grade 3)
2.OA.B.2Fluently add and subtract within $20$ (Computing the small running totals $S_1 = 1, S_2 = 4, S_3 = 9, S_4 = 16$ by hand to generate data.)3.OA.D.9Identify arithmetic patterns and explain them using properties of operations (Spotting that the running totals $1, 4, 9, 16, 25$ are the perfect squares $n^2$ and confirming the rule with one extra case.)3.OA.C.7Fluently multiply and divide within $100$ (Computing $20^2 = 400$ as the final running total after $20$ days.)
⭐ This AMC 8 problem only needs the Grade 3 fact that the sum of the first $n$ odd numbers is $n^2$ — which you can discover yourself with Grade 2 addition!
⭐ This AMC 8 problem only needs the Grade 3 fact that the sum of the first $n$ odd numbers is $n^2$ — which you can discover yourself with Grade 2 addition!
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