AMC 10 · 2021 · #8

Grade 5 geometry-2d

Archetype Lattice / Grid Pattern from Small Cases └ Sub-archetype Coordinate Spiral Position

pattern-recognitionperfect-squaresspatial-visualizationsystematic-enumeration pattern-recognitioneasier-related-problem ↑ Prerequisites: perfect-squarespattern-recognition

📏 Medium solution 💡 3 insights 📊 Diagram

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Problem

Mr. Zhou writes 1 in the middle cell (row 8, column 8) of a 15 × 15 grid and then writes 2, 3, 4, …, 225 in a clockwise spiral outward — first one step east, then turning to fill each new ring. Look at the second row from the top of the grid. Find the sum of the largest number and the smallest number that appear in that row.

Pick an answer.

(A)

~367

(B)

~368

(C)

~369

(D)

~379

(E)

~380

AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Easier Related Problem

Tool #9 (Easier Related Problem): the given diagram already solves the 7 × 7 sub-version — use it to learn the spiral rule, then extend to 15 × 15. Tool #5 (Pattern): the top-right corner of ring k is the odd square (2k-1)², and each ring adds 8k cells along 1+ (2k-1) + 2k + 2k + 2k east/south/west/north/east moves. Tool #1 (Diagram): sketch the outer two rings (ring 7 and ring 8) to locate row 2. Tool #7 (Subproblems): split row 2 into 'inside ring 8' (just the two endpoints) and 'inside ring 7 top edge' (the middle 13 cells), then take max and min separately.

1STEP 1

The spiral's top-right corners are the odd squares 1, 9, 25, …, so ring k's corner is (2k-1)² (ring 8 ends at 15²=225).

top-right of ring k = (2k-1)²

2STEP 2

The 7×7 picture shows the moves corner-to-corner: 1 east, then south, west, north, east — each new ring adds 8k cells.

(2k+1)² - (2k-1)² = 8k ✓

3STEP 3

Ring 7's corner 13²=169 sits at row 2, column 14; the east-step before it fills row 2 columns 3–14 with 158, 159, …, 169.

row 2, cols 3 -- 14 → 158, 159, …, 169

4STEP 4

The north-step up ring 7's left edge ends at row 2, column 2 with value 157, so row 2 across ring 7 reads 157, 158, …, 169.

row 2, col 2 = 157; row 2, cols 2 -- 14 → 157, 158, …, 169

5STEP 5

Ring 8 starts at 170: the east-step puts 170 at row 2 col 15; the north-step (ending 211 at row 1) puts 210 at row 2 col 1.

row 2, col 15 = 170; row 2, col 1 = 210

6STEP 6

Row 2 reads 210, 157, 158, …, 169, 170 — largest 210, smallest 157, so max+min = 210+157 = 367, choice (A).

max + min = 210 + 157 = 367 → (A)

Answer

~367

Sanity: every value in row 2 should lie in {157, 158, …, 170} ∪ {210} — and indeed all these are between 1 and 225. The extremes are min = 157 and max = 210, and 367 is in the right ballpark (between 2 · 150 = 300 and 2 · 225 = 450). Among the choices, only (A) 367 matches a clean integer split into a middle-band value plus an outer-ring value.

💡Key takeaway

This AMC 10 problem only needs Grade 5 number patterns you already know — the spiral's corners are the odd squares (2k-1)², so ring 7 ends at 169 on row 2 column 14, the ring's top edge runs from 157 to 169, and ring 8's neighbors put 170 at column 15 and 210 at column 1. Max + min = 210 + 157 = 367, choice (A).

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