AMC 10 · 2012 · #14
Easy mode Grade 4Chubby makes a checkerboard that is 31 squares across and 31 squares down. Every corner square is black. Along each row and each column, the squares switch back and forth between black and red. How many black squares are on the board?
Pick an answer.
AMC 10 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: The board is a $31\times31$ grid of squares.; All four corner squares are black.; Colors alternate red and black along every row and every column.; Answer choices: (A) $480$, (B) $481$, (C) $482$, (D) $483$, (E) $484$.
Unknowns: The total number of black squares on the $31\times31$ board.
Understand
Restated: 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.
Givens: The board is a $31\times31$ grid of squares.; All four corner squares are black.; Colors alternate red and black along every row and every column.; Answer choices: (A) $480$, (B) $481$, (C) $482$, (D) $483$, (E) $484$.
Plan
Primary tool: #5 Look for a Pattern
Secondary: #1 Draw a Diagram, #7 Identify Subproblems
The full $31\times31$ 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.
Execute — Answer: B
2.OA.C.4 Step 1 Sketch a small odd board
- Draw a tiny odd-sided board, say $3\times3$, with black in every corner.
- Reading any row you get black, red, black — the strip starts and ends black because its length is odd.
- The same thing happens on the $31$-wide board: because $31$ is odd, every row starts and ends with the same color as its two end squares.
💡 An odd-length strip always ends with the same color it began with.
4.OA.C.5 Step 2 Count black squares per row type
- A row of $31$ squares that starts and ends black looks like black, red, black, red, ...
- , black.
- The blacks and reds alternate, and since it has one more black than red, the $31$ squares split into $16$ black and $15$ red.
- A row that starts and ends red is the mirror image: $15$ black and $16$ red.
- So every row holds either $16$ black squares or $15$ black squares.
💡 In an alternating strip of odd length, the color at the ends appears exactly once more than the other.
2.OA.C.3 Step 3 Count how many rows of each type
- The left column also alternates and starts black at the top corner, so the rows that begin black are rows $1,3,5,\ldots,31$ and the rows that begin red are rows $2,4,6,\ldots,30$.
- Among the numbers $1$ to $31$ there are $16$ odd row numbers and $15$ even row numbers.
- So $16$ rows are black-ended and $15$ rows are red-ended.
💡 The corner color marks the odd rows; counting odd versus even numbers up to $31$ splits the rows.
4.NBT.B.5 Step 4 Multiply each group
- Multiply the count of each row type by how many black squares that type contributes.
- The $16$ black-ended rows give $16\times16=256$ black squares, and the $15$ red-ended rows give $15\times15=225$ black squares.
💡 Same-size groups turn a long count into two quick multiplications.
4.NBT.B.4 Step 5 Add the two groups
- Add the black squares from both groups: $256+225=481$.
- That is the total number of black squares on the board, which is choice (B).
- The nearby choices come from small slips: $480$ if you give the extra square to red, and $482$ or more if you miscount a row.
💡 The two row groups don't overlap, so their black counts simply add.
2.OA.C.4 Draw a tiny odd-sided board, say $3\times3$, with black in every corner. Reading 4.OA.C.5 A row of $31$ squares that starts and ends black looks like black, red, black, r 2.OA.C.3 The left column also alternates and starts black at the top corner, so the rows 4.NBT.B.5 Multiply the count of each row type by how many black squares that type contribu 4.NBT.B.4 Add the black squares from both groups: $256+225=481$. That is the total number Review
Reasonableness: The board has $31^2=961$ 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 $\dfrac{961+1}{2}=481$ black squares. This matches the row-by-row count. The trap answer $480$ is exactly $\dfrac{961-1}{2}$, what you get if you wrongly hand the extra square to red.
Alternative: Skip the rows entirely and use parity: on any odd-by-odd board, the color in the corners takes $\dfrac{n^2+1}{2}$ squares. With $n=31$ that is $\dfrac{961+1}{2}=481$ in a single step.
CCSS standards used (min grade 4)
2.OA.C.4Use addition to find the total number of objects in rectangular arrays (Seeing the checkerboard as $31$ rows of $31$ squares and reading each row as an ordered strip.)4.OA.C.5Generate a number or shape pattern following a given rule (Using the alternating color rule to find that an odd-length row holds $16$ of one color and $15$ of the other.)2.OA.C.3Determine whether a group of objects has an odd or even number (Counting $16$ odd-numbered (black-ended) rows and $15$ even-numbered (red-ended) rows among rows $1$ to $31$.)4.NBT.B.5Multiply a whole number of up to four digits by a one-digit whole number (Computing $16\times16=256$ and $15\times15=225$ for the two groups of rows.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Adding $256+225=481$ to combine the black squares from both row groups.)
⭐ On an odd-by-odd board, whichever color sits in the corners gets exactly one more than half of all the squares.
⭐ On an odd-by-odd board, whichever color sits in the corners gets exactly one more than half of all the squares.
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