On her first day of work, Janabel sold one widget. On day two, she sold three widgets. On day three, she sold five widgets, and on each succeeding day, she sold two more widgets than she had sold on the previous day. How many widgets in total had Janabel sold after working 20 days?
Try it yourself first — the explanation is most useful after you’ve attempted it.
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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
#9 Solve an Easier Related Problem 2.OA.B.2Step 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.
#5 Look for a Pattern 3.OA.D.9Step 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 $n^2$.
$$S_n = n^2 \text{ for } n = 1, 2, 3, 4$$
💡 Spotting a numerical pattern in the running totals is the Grade 3 "identify arithmetic patterns" standard in action.
#5 Look for a Pattern 3.OA.D.9Step 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.
$$S_5 = 16 + 9 = 25 = 5^2 \;\checkmark$$
💡 A pattern from $4$ cases can still fool you — testing one more case is the cheap insurance the toolkit asks for.
#5 Look for a Pattern 3.OA.C.7Step 4
Apply the pattern to $n = 20$.
The total widgets sold after $20$ days is $20^2$.
$$S_{20} = 20^2 = 400$$
💡 Squaring a multiple of $10$ is a Grade 3 single-digit multiplication fact ($2 \times 2 = 4$) with two extra zeros tacked on.
#3 Eliminate Possibilities 3.OA.C.7Step 5
Match to the answer choices.
$400$ appears as option (D); none of the other choices are square numbers of $20$.
$$400 \;\Rightarrow\; \textbf{(D)}$$
💡 Eliminating answer choices is the standard AMC multiple-choice safety check.
[1]
#9 2.OA.B.2Compute the running total for the first few days. This is the easier version of
[2]
#5 3.OA.D.9Compare the running totals to a familiar pattern. The list $1, 4, 9, 16$ is exac
[3]
#5 3.OA.D.9Check the pattern on one more case so we trust it before jumping to $n = 20$. Da
[4]
#5 3.OA.C.7Apply the pattern to $n = 20$. The total widgets sold after $20$ days is $20^2$.
[5]
#3 3.OA.C.7Match 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.2 Fluently 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.9 Identify 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.7 Fluently 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!