AMC 8 · 2002 · #11

Grade 4 geometry-2dcounting

Archetype Lattice / Grid Pattern from Small Cases └ Sub-archetype Figure-N Growth Sequence

perfect-squarespattern-recognitionsequences-arithmetic pattern-recognitionidentify-subproblems ↑ Prerequisites: multi-digit-arithmeticperfect-squares

📏 Short solution 💡 2 insights 📊 Diagram

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Problem

A sequence of squares is made of identical square tiles. The edge of each square is one tile length longer than the edge of the previous square. The first three squares are shown. How many more tiles does the seventh square require than the sixth?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2002 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: A sequence of squares is built from unit tiles. Each square's side is one tile longer than the previous square's side. How many more tiles does the $7$th square use than the $6$th square?

Givens: The squares are made of identical unit tiles, with no gaps; Each square's edge is one tile longer than the previous square's edge; From the figure the first three squares have edges $1$, $2$, $3$; Answer choices: (A) $11$, (B) $12$, (C) $13$, (D) $14$, (E) $15$

Unknowns: $(\text{tiles in the }7\text{th square}) - (\text{tiles in the }6\text{th square})$

Understand

Plan

Primary tool: #5 Look for a Pattern

Secondary: #9 Solve an Easier Related Problem

The picture already gives the first three cases for free, which is the cue for Tool #5 (Look for a Pattern): count the tiles in those small squares and see what rule governs them. Tool #9 (Solve an Easier Related Problem) is what makes Tool #5 honest — we trust the pattern only after verifying it on the $n=1, 2, 3$ cases we can see and count. Once the rule "$n$-th square has $n^2$ tiles" is locked in, the $6$th and $7$th squares come from one subtraction. No algebra needed.

Execute — Answer: C

#9 Solve an Easier Related Problem 3.MD.C.7 Step 1

Count tiles in the small cases shown in the figure.
The first square is $1 \times 1$, the second is $2 \times 2$, the third is $3 \times 3$.

$$1\text{st}: 1 \text{ tile}, \quad 2\text{nd}: 4 \text{ tiles}, \quad 3\text{rd}: 9 \text{ tiles}$$

💡 Counting unit tiles in an $n$-by-$n$ square is exactly the Grade 3 area-by-tiling idea: side $\times$ side.

#5 Look for a Pattern 4.OA.C.5 Step 2

Look at the counts $1, 4, 9$ and name the pattern.
They are $1^2, 2^2, 3^2$, so the $n$-th square uses $n^2$ tiles.
Verify the rule on the next case: the $4$th square would be $4 \times 4 = 16$ tiles, which matches adding one more row and column.

$$\text{tiles in } n\text{-th square} = n^2$$

💡 Three matching cases plus a confirmed fourth is enough to trust the rule — that is the Grade 4 "generate and analyze a pattern" move.

#5 Look for a Pattern 3.MD.C.7 Step 3

Apply the rule to the $6$th and $7$th squares.

$$6\text{th}: 6^2 = 36, \quad 7\text{th}: 7^2 = 49$$

💡 Same area-as-tile-count picture, just bigger sides.

#5 Look for a Pattern 3.OA.D.8 Step 4

Subtract to get the extra tiles needed.

$$49 - 36 = 13 \;\Rightarrow\; \textbf{(C)}$$

💡 The question asks "how many more," which is a Grade 3 take-away comparison.

[1] #9 3.MD.C.7 Count tiles in the small cases shown in the figure. The first square is $1 \time

[2] #5 4.OA.C.5 Look at the counts $1, 4, 9$ and name the pattern. They are $1^2, 2^2, 3^2$, so

[3] #5 3.MD.C.7 Apply the rule to the $6$th and $7$th squares.

[4] #5 3.OA.D.8 Subtract to get the extra tiles needed.

Review

Reasonableness: Sanity check by building the $7$th square from the $6$th. Start with the $6 \times 6$ square ($36$ tiles). To grow it into a $7 \times 7$, add one new row of $7$ tiles along the top and one new column of $7$ tiles along the right — but the top-right corner tile is shared. Extra tiles $= 7 + 7 - 1 = 13$, which matches $49 - 36 = 13$. The differences between consecutive square counts are $4-1=3, 9-4=5, 16-9=7, \ldots$, the odd numbers, so the gap between the $6$th and $7$th squares is the $6$th odd number $= 2 \cdot 6 + 1 = 13$. Same answer (C).

Alternative: Tool #1 (Draw a Diagram): on top of the $6 \times 6$ grid, draw the L-shaped border that turns it into a $7 \times 7$. The L is one row of $7$ on top plus one column of $7$ on the right, with the corner counted once: $7 + 7 - 1 = 13$ tiles. The diagram makes the answer (C) without any squaring.

CCSS standards used (min grade 4)

3.MD.C.7 Relate area to multiplication, finding the area of a rectangle by tiling and by side $\times$ side (Counting tiles in each $n$-by-$n$ square as $n \times n$: $1, 4, 9$ for the first three squares and $36, 49$ for the sixth and seventh.)
3.OA.D.8 Solve two-step word problems using the four operations (Computing the "how many more" comparison $49 - 36 = 13$ at the end.)
4.OA.C.5 Generate and analyze a number or shape pattern that follows a given rule (Reading $1, 4, 9$ as $1^2, 2^2, 3^2$ and extending the rule "$n$-th square has $n^2$ tiles" to $n = 6$ and $n = 7$.)

⭐ Count the small cases ($1, 4, 9$), spot the square-number rule, then subtract: $7^2 - 6^2 = 49 - 36 = 13$. Answer (C).

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 4)

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