AMC 8 · 2009 · #18
Grade 4 patternProblem
The diagram represents a 7-foot-by-7-foot floor that is tiled with 1-square-foot black tiles and white tiles. Notice that the corners have white tiles. If a 15-foot-by-15-foot floor is to be tiled in the same manner, how many white tiles will be needed?
Pick an answer.
AMC 8 2009 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: A $7$-by-$7$ floor is tiled with $1$-square-foot black and white tiles in a striped pattern: the corner tiles are white, and black stripes (rows and columns) sit one position in from each edge and continue every other position. The same rule is applied to a $15$-by-$15$ floor. Count the white tiles needed for the $15$-by-$15$ floor.
Givens: Each tile is $1$ square foot; The $7$-by-$7$ picture shows black tiles forming full rows at positions $2, 4, 6$ and full columns at positions $2, 4, 6$ (counting from one edge); All four corner tiles are white; The new floor is $15$ by $15$, tiled by the same rule; Answer choices: (A) $49$, (B) $57$, (C) $64$, (D) $96$, (E) $126$
Unknowns: The number of white tiles in the $15$-by-$15$ floor
Understand
Restated: A $7$-by-$7$ floor is tiled with $1$-square-foot black and white tiles in a striped pattern: the corner tiles are white, and black stripes (rows and columns) sit one position in from each edge and continue every other position. The same rule is applied to a $15$-by-$15$ floor. Count the white tiles needed for the $15$-by-$15$ floor.
Givens: Each tile is $1$ square foot; The $7$-by-$7$ picture shows black tiles forming full rows at positions $2, 4, 6$ and full columns at positions $2, 4, 6$ (counting from one edge); All four corner tiles are white; The new floor is $15$ by $15$, tiled by the same rule; Answer choices: (A) $49$, (B) $57$, (C) $64$, (D) $96$, (E) $126$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #9 Solve an Easier Related Problem
Jumping straight to a $15$-by-$15$ count is messy. Tool #9 (Easier Related Problem) says: try the same tiling on the smallest odd-side floors first — $1 \times 1$, $3 \times 3$, $5 \times 5$, $7 \times 7$ — and count whites by hand. Tool #5 (Look for a Pattern) then turns the four counts into a rule we can apply to side $15$. The picture already does the $7 \times 7$ case for us, so most of the work is just counting and matching.
Execute — Answer: C
2.G.A.2 Step 1 - Look at where the white tiles sit on the $7 \times 7$ picture.
- The black stripes are the rows and columns at positions $2, 4, 6$, so white tiles are exactly the squares whose row AND column are in $\{1, 3, 5, 7\}$ — the odd positions.
- That's $4$ odd rows and $4$ odd columns.
💡 Partitioning a square into rows and columns of unit tiles and counting an intersection pattern is a Grade 2 array idea.
2.G.A.2 Step 2 - Repeat the same counting for the smaller odd-side floors.
- On a $1 \times 1$ floor the single tile is a corner, so it is white: $1$ white.
- On $3 \times 3$ the odd-position rows and columns are $\{1, 3\}$, giving $2 \times 2 = 4$ whites.
- On $5 \times 5$ the odd positions are $\{1, 3, 5\}$, giving $3 \times 3 = 9$ whites.
💡 Smaller, hand-countable versions of the same tiling let us see the structure without algebra.
4.OA.C.5 Step 3 - The four counts $1, 4, 9, 16$ are exactly the perfect squares $1^2, 2^2, 3^2, 4^2$.
- The side lengths $1, 3, 5, 7$ are the $1$st, $2$nd, $3$rd, $4$th odd numbers.
- So for the $k$-th odd-side floor, the number of white tiles is $k^2$.
💡 Generating a square-number pattern from a tiling rule is exactly the Grade 4 "generate a number or shape pattern" standard.
3.OA.D.9 Step 4 - Find which odd number $15$ is.
- The odd numbers go $1, 3, 5, 7, 9, 11, 13, 15$, so $15$ is the $8$th odd number, meaning $k = 8$.
💡 Recognizing odd numbers as $1, 3, 5, \dots$ and locating $15$ in the sequence is a Grade 3 arithmetic-pattern move.
3.OA.C.7 Step 5 Apply the pattern at $k = 8$.
💡 Computing $8 \times 8 = 64$ is exactly the Grade 3 "multiply fluently within $100$" standard.
2.G.A.2 Look at where the white tiles sit on the $7 \times 7$ picture. The black stripes 2.G.A.2 Repeat the same counting for the smaller odd-side floors. On a $1 \times 1$ floo 4.OA.C.5 The four counts $1, 4, 9, 16$ are exactly the perfect squares $1^2, 2^2, 3^2, 4^ 3.OA.D.9 Find which odd number $15$ is. The odd numbers go $1, 3, 5, 7, 9, 11, 13, 15$, s 3.OA.C.7 Apply the pattern at $k = 8$. Review
Reasonableness: Sanity check the pattern from a different angle. On any odd-side floor of size $s = 2k - 1$, the white tiles form a $k$-by-$k$ grid (one white tile at every odd-row, odd-column intersection), so the white count is $k^2$. Plug in $s = 7$: $k = 4$, $k^2 = 16$ — which matches the picture. Plug in $s = 15$: $k = 8$, $k^2 = 64$. The answer $64$ is also well below the $15 \times 15 = 225$ total tiles, which is the right ballpark since black stripes cover most of the floor.
Alternative: Tool #7 (Identify Subproblems) — count by complement. The $15 \times 15$ has $225$ tiles total. Black rows sit at positions $2, 4, 6, 8, 10, 12, 14$ ($7$ rows of $15$ = $105$ tiles). Black columns sit at the same positions ($7$ columns of $15$ = $105$ tiles). The overlap (cells in a black row AND a black column) is $7 \times 7 = 49$ tiles, counted twice. So black $= 105 + 105 - 49 = 161$, and white $= 225 - 161 = 64$. Same answer.
CCSS standards used (min grade 4)
2.G.A.2Partition a rectangle into rows and columns of same-size squares (Counting white tiles on the $7 \times 7$ picture and the smaller $3 \times 3$, $5 \times 5$ floors by reading off odd-position rows and columns.)3.OA.C.7Fluently multiply and divide within 100 (Computing $8 \times 8 = 64$ once the pattern $k^2$ is known.)3.OA.D.9Identify arithmetic patterns and explain using properties of operations (Identifying $1, 3, 5, 7, \dots$ as the odd-number sequence and locating $15$ as the $8$th odd number.)4.OA.C.5Generate a number or shape pattern following a given rule (Generalizing the $1, 4, 9, 16$ counts as $k^2$ for the $k$-th odd-side floor.)
⭐ This AMC 8 problem only needs Grade 4 pattern-finding with square numbers you already know!
⭐ This AMC 8 problem only needs Grade 4 pattern-finding with square numbers you already know!
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