AMC 8 · 2014 · #6

Grade 4 arithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Fraction / Expression Sum Evaluation

area-rectanglesperfect-squaresmulti-digit-arithmetic identify-subproblems ↑ Prerequisites: area-rectanglesmulti-digit-arithmetic

📏 Short solution 💡 2 insights

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Problem

Six rectangles each with a common base width of $2$ have lengths of $1, 4, 9, 16, 25$ , and $36$ . What is the sum of the areas of the six rectangles?

Pick an answer.

(A)

(B)

(C)

162

(D)

182

(E)

202

AMC 8 2014 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.

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Toolkit + CCSS Solution

Understand

Restated: Six rectangles each have width $2$. Their lengths are the first six perfect squares: $1, 4, 9, 16, 25, 36$. Find the sum of their areas.

Givens: There are $6$ rectangles; Every rectangle has the same width: $2$; Lengths are $1, 4, 9, 16, 25, 36$ (the first six perfect squares); Answer choices: (A) $91$, (B) $93$, (C) $162$, (D) $182$, (E) $202$

Unknowns: The total area of all six rectangles combined

Understand

Restated: Six rectangles each have width $2$. Their lengths are the first six perfect squares: $1, 4, 9, 16, 25, 36$. Find the sum of their areas.

Plan

Primary tool: #4 Find a Pattern

Secondary: #7 Identify Subproblems

Each rectangle's area is $2 \times \text{length}$, so the total area is $2 \times 1 + 2 \times 4 + \dots + 2 \times 36$. The shared factor $2$ is a pattern (Tool #4) — every term has it — so we can factor it out and add the lengths just once instead of six times: $\text{total} = 2 \times (1 + 4 + 9 + 16 + 25 + 36)$. Tool #7 (Identify Subproblems) then splits the work into two clean pieces — first add the six numbers, then multiply by $2$.

Execute — Answer: D

#7 Identify Subproblems 3.MD.C.7 Step 1

Write the total area as a sum of six rectangle areas.
Each rectangle area is width $\times$ length $= 2 \times \ell$.

$$\text{Total} = 2\cdot1 + 2\cdot4 + 2\cdot9 + 2\cdot16 + 2\cdot25 + 2\cdot36$$

💡 Area $=$ width $\times$ length for a rectangle is the Grade 3 area standard.

#4 Find a Pattern 3.OA.B.5 Step 2

Notice the shared factor of $2$ in every term.
Pull it out using the distributive property so we only have to add the lengths once.

$$\text{Total} = 2 \times (1 + 4 + 9 + 16 + 25 + 36)$$

💡 Factoring out a common factor is exactly the distributive property of multiplication over addition.

#7 Identify Subproblems 4.NBT.B.4 Step 3

Add the six perfect squares inside the parentheses.
Group them in easy pairs to keep the arithmetic clean: $(1 + 9) + (4 + 16) + (25 + 36) = 10 + 20 + 61$.

$$1 + 4 + 9 + 16 + 25 + 36 = 10 + 20 + 61 = 91$$

💡 Pairing numbers that add to round values (like $10$ and $20$) is a standard mental-math regrouping move.

#7 Identify Subproblems 4.NBT.B.5 Step 4

Multiply the sum by the common width $2$ to get the total area.

$$\text{Total} = 2 \times 91 = 182 \;\Rightarrow\; \textbf{(D)}$$

💡 A one-digit by two-digit multiplication is Grade 4 multi-digit arithmetic.

[1] #7 3.MD.C.7 Write the total area as a sum of six rectangle areas. Each rectangle area is wid

[2] #4 3.OA.B.5 Notice the shared factor of $2$ in every term. Pull it out using the distributiv

[3] #7 4.NBT.B.4 Add the six perfect squares inside the parentheses. Group them in easy pairs to

[4] #7 4.NBT.B.5 Multiply the sum by the common width $2$ to get the total area.

Review

Reasonableness: Choice (A) $91$ is the sum of the lengths alone, before multiplying by the width $2$ — a tempting trap if you forget the last step. Doubling $91$ gives $182$, which is choice (D). Quick sanity check: the largest single rectangle is $2 \times 36 = 72$, and the six areas $2, 8, 18, 32, 50, 72$ are all well under $200$, so a total near $182$ is reasonable.

Alternative: Tool #1 (Make a List): compute each area separately — $2, 8, 18, 32, 50, 72$ — and add: $2 + 8 = 10$, $10 + 18 = 28$, $28 + 32 = 60$, $60 + 50 = 110$, $110 + 72 = 182$. Same answer (D), but with six multiplications instead of one.

CCSS standards used (min grade 4)

3.MD.C.7 Relate area to the operations of multiplication and addition (Using area $=$ width $\times$ length for each rectangle to write the total as a sum of six products.)
3.OA.B.5 Apply properties of operations as strategies to multiply and divide (Factoring out the common width $2$ via the distributive property: $2\cdot1 + 2\cdot4 + \dots = 2 \times (1 + 4 + \dots)$.)
4.NBT.B.4 Fluently add and subtract multi-digit whole numbers (Adding the six perfect squares $1 + 4 + 9 + 16 + 25 + 36 = 91$.)
4.NBT.B.5 Multiply a whole number of up to four digits by a one-digit whole number (Computing the final total $2 \times 91 = 182$.)

⭐ This AMC 8 problem only needs Grade 4 arithmetic — once you factor out the shared width, it's just one addition and one multiplication!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 4)

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