AMC 8 · 2017 · #11

Grade 4 geometry-2dpattern

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

perfect-squaresparitypattern-recognitionspatial-visualization pattern-recognitioneasier-related-problem ↑ Prerequisites: multi-digit-arithmeticparity

📏 Medium solution 💡 3 insights

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Problem

A square floor is paved with congruent square tiles, forming an n × n grid. The tiles sitting on the two main diagonals of the floor total 37. Find the total number of tiles covering the whole floor.

Pick an answer.

(A)

148

(B)

324

(C)

361

(D)

1296

(E)

1369

AMC 8 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Solve an Easier Related Problem

We do not know the side length n, and trying to attack n² directly with the value 37 is awkward. Tool #9 (Easier Related Problem) says: shrink the floor first — draw a 2 × 2, 3 × 3, 4 × 4, 5 × 5 grid (Tool #1) and count the diagonal tiles in each. Tool #5 (Look for a Pattern) then reveals the rule: odd n gives 2n - 1 diagonal tiles, even n gives 2n. Because 37 is odd, n must be odd, and 2n - 1 = 37 pinpoints n. We finish by computing n².

1STEP 1

Draw tiny grids and count diagonal tiles: even sides share none, odd sides share the center — giving 4, 5, 8, 9 for n = 2, 3, 4, 5.

n=2: 2+2=4 n=3: 3+3-1=5 n=4: 4+4=8 n=5: 5+5-1=9

2STEP 2

Line up the counts: even n gives 2n, odd n gives 2n - 1 — so the total's parity matches n's parity.

diagonal tiles = 2n & n even ; 2n - 1 & n odd

3STEP 3

37 is odd, so n is odd: solve 2n - 1 = 37 to get n = 19.

2n - 1 = 37 → 2n = 38 → n = 19

4STEP 4

Total tiles = area of the 19 × 19 grid: (20 - 1)² = 400 - 40 + 1 = 361, choice (C).

n² = 19 × 19 = 361 → (C)

Answer

361

Sanity check the answer choices first. (A) 148 is not a perfect square, so it cannot be n² for an integer n — eliminate. (D) 1296 = 36²: with n = 36 (even) the diagonals would give 2 × 36 = 72 tiles, not 37 — eliminate. (E) 1369 = 37²: with n = 37 (odd) the diagonals would give 2 × 37 - 1 = 73 tiles, not 37 — eliminate. (B) 324 = 18²: n = 18 is even, giving 36 diagonal tiles, not 37 — eliminate. (C) 361 = 19²: n = 19 is odd, giving 2 × 19 - 1 = 37 diagonal tiles. Matches.

💡Key takeaway

This AMC 8 problem only needs Grade 4 pattern-finding and two-digit multiplication you already know!

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