2012 AMC 10A Problem 14

P11-15~3 minPrint

Paths & Grids Number Properties Set up the equation/formula and compute; no special trick needed Use an invariant, parity, or coloring argument

Chubby makes nonstandard checkerboards that have $31$ squares on each side. The checkerboards have a black square in every corner and alternate red and black squares along every row and column. How many black squares are there on such a checkerboard?

Show solution without answering

Key insight

A row of 31 starting black has 16 black squares; one starting red has 15. The 31 rows alternate, starting black, so 16 rows have 16 blacks.

Solution

1. Look at a single row first. It has $31$ squares, alternating colors. If it starts with black, the blacks are in spots $1, 3, 5, \dots, 31$ , which is $16$ squares. If it starts with red, the blacks are in spots $2, 4, \dots, 30$ , which is $15$ squares.
2. Which rows start with black? The first square of each row is in the first column, and that column also alternates. The top corner is black, so rows $1, 3, 5, \dots, 31$ start with black and rows $2, 4, \dots, 30$ start with red.
3. That makes $16$ rows starting with black and $15$ rows starting with red.
4. Count the black squares:
$16 \times 16 + 15 \times 15 = 256 + 225 = 481.$
5. Check the corners: row $31$ starts with black and ends with black (spot $31$ is odd), so all four corners are black, as the problem says.

The answer is $\boxed{\textbf{(B)}\ 481}$ .

Why this works

On a board with an even number of squares per side, the colors split exactly in half. With an odd side, the color in the corners gets one extra square: here $961$ squares split as $481$ black and $480$ red. A small case confirms the pattern: a $3 \times 3$ board with black corners has $5$ black and $4$ red squares.

Alternative approach

Pair up squares within each row. A row of $31$ has $15$ pairs of red and black, plus one leftover square of the color the row starts with. The $15$ pairs in each of the $31$ rows give $15 \times 31 = 465$ black squares. The leftovers are black in the $16$ rows that start black. Total: $465 + 16 = 481$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Number the rows and columns $1$ through $31$ . Alternating colors along every row and column means the color of square $(r, c)$ depends only on whether $r + c$ is even or odd. The corners $(1,1)$ , $(1,31)$ , $(31,1)$ , $(31,31)$ all have even sums and are black, so black means $r + c$ even.

Count squares with $r + c$ even. Row $1$ (odd) has black squares in the $16$ odd columns; row $2$ (even) has black squares in the $15$ even columns. The pattern alternates $16, 15, 16, 15, \dots$ over the $31$ rows: $16$ odd rows contribute $16$ each and $15$ even rows contribute $15$ each:
$16 \cdot 16 + 15 \cdot 15 = 256 + 225 = 481.$

The answer is $\boxed{\textbf{(B)}\ 481}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 5: More With Combinations (§5.2)
Introduction to Number Theory, Chapter 1: Integers: The Basics
Introduction to Algebra, Chapter 20: Special Functions (§20.3)