AMC 8 · 2015 · #9

Easy mode Grade 3

Problem

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?

$\textbf{(A) }39\qquad\textbf{(B) }40\qquad\textbf{(C) }210\qquad\textbf{(D) }400\qquad \textbf{(E) }401$

Pick an answer.

(A)

(B)

(C)

210

(D)

400

(E)

401

View mode:

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

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.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 $n = 20$).

$$S_1 = 1,\; S_2 = 1+3 = 4,\; S_3 = 1+3+5 = 9,\; S_4 = 1+3+5+7 = 16$$

💡 Adding the first few odd numbers is just Grade 2 addition within $20$ — small, safe, and it gives us real data to look at.

#5 Look for a Pattern 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 $n^2$.

$$S_n = n^2 \text{ for } n = 1, 2, 3, 4$$

💡 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.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.

$$S_5 = 16 + 9 = 25 = 5^2 \;\checkmark$$

💡 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.7 Step 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.7 Step 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.2 Compute the running total for the first few days. This is the easier version of

[2] #5 3.OA.D.9 Compare the running totals to a familiar pattern. The list $1, 4, 9, 16$ is exac

[3] #5 3.OA.D.9 Check the pattern on one more case so we trust it before jumping to $n = 20$. Da

[4] #5 3.OA.C.7 Apply the pattern to $n = 20$. The total widgets sold after $20$ days is $20^2$.

[5] #3 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.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!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 3)

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