AMC 8 · 2009 · #18

Grade 4 pattern
Archetype Lattice / Grid Pattern from Small Cases └ Sub-archetype Figure-N Growth Sequence
pattern-recognitionperfect-squaresarea-rectangles pattern-recognitioneasier-related-problem ↑ Prerequisites: area-rectanglesperfect-squares
📏 Medium solution 💡 3 insights 📊 Diagram
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Problem
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.

Pick an answer.

(A)
49
(B)
57
(C)
64
(D)
96
(E)
126

AMC 8 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Look for a Pattern

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 × 1, 3 × 3, 5 × 5, 7 × 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 × 7 case for us, so most of the work is just counting and matching.

1STEP 1

A tile is white only where both its row and column are odd, so the 7 × 7 has 4 odd rows times 4 odd columns = 16 whites.

white tiles in 7× 7 = 4 × 4 = 16
2STEP 2

Counting the same way, 1 × 1 gives 1 white, 3 × 3 gives 4, and 5 × 5 gives 9 — joining 16 to make the list 1, 4, 9, 16.

1, 4, 9, 16
3STEP 3

Those counts are the squares 1², 2², 3², 4², so a floor whose side is the k-th odd number has k² white tiles.

whites(s) = k² where s = 2k - 1
4STEP 4

The odd numbers 1, 3, 5, 7, 9, 11, 13, 15 put 15 in eighth place, so k = 8.

15 = 2 × 8 - 1 → k = 8
5STEP 5

Applying the rule at k = 8 gives 8² = 64 white tiles — choice (C).

whites(15) = 8² = 64 → (C)
Answer
64
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 k². Plug in s = 7: k = 4, k² = 16 — which matches the picture. Plug in s = 15: k = 8, k² = 64. The answer 64 is also well below the 15 × 15 = 225 total tiles, which is the right ballpark since black stripes cover most of the floor.
💡Key takeaway

This AMC 8 problem only needs Grade 4 pattern-finding with square numbers you already know!

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