AMC 10 · 2012 · #14

Grade 4 counting
Archetype Lattice / Grid Pattern from Small Cases └ Sub-archetype Periodic Placement Formula
pattern-recognitionparity parity-coloringsystematic-enumeration ↑ Prerequisites: multi-digit-arithmetic
📏 Medium solution 💡 2 insights
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Problem
A checkerboard is 31 squares wide and 31 squares tall. Every corner square is black, and the colors switch between black and red along each row and each column. Count how many black squares the whole board has.

Pick an answer.

(A)
480
(B)
481
(C)
482
(D)
483
(E)
484

AMC 10 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Look for a Pattern

The full 31×31 board is too big to count square by square, so Tool #5 (Look for a Pattern) turns it into a repeating rule: each row is just an alternating strip, and the number of black squares in a row depends only on the color at its ends. Tool #1 (Draw a Diagram) grounds that rule on a small odd board first, and Tool #7 (Identify Subproblems) splits the 31 rows into two clean groups — black-ended rows and red-ended rows — that are easy to count and add.

1STEP 1

Sketch a small odd board

Sketch a 3×3 board with black corners: a row reads black, red, black — odd length ends in the color it started.

3×3 row: ■ □ ■
2STEP 2

Count black squares per row type

A 31-square row with black ends splits into 16 black and 15 red; a red-ended row is the mirror, only 15 black.

black-ended row=16 black, red-ended row=15 black
3STEP 3

Count how many rows of each type

The left column alternates down from a black corner, so odd-numbered rows are black-ended — 16 such rows and 15 red-ended ones.

16 black-ended rows, 15 red-ended rows
4STEP 4

Multiply each group

Multiply each group by its own black count: the 16 black-ended rows give 256, the 15 red-ended rows give 225.

16×16=256, 15×15=225
5STEP 5

Add the two groups

Add the two groups: 256+225=481 black squares, choice (B); 480 is what a red-favoring slip gives.

256+225=481 → (B)
Answer
481
The board has 31²=961 squares in all. That total is odd, so black and red cannot split it evenly — one color must get the extra square. Since all four corners are black, black is the color that wins the extra, giving (961+1)/2=481 black squares. This matches the row-by-row count. The trap answer 480 is exactly (961-1)/2, what you get if you wrongly hand the extra square to red.
💡Key takeaway

On an odd-by-odd board, whichever color sits in the corners gets exactly one more than half of all the squares.

  • Sketch a small odd board
  • Count black squares per row type
  • Count how many rows of each type
  • Multiply each group
  • Add the two groups

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