AMC 10 · 2004 · #16
Grade 3 geometry-2dThe 5×5 grid shown contains a collection of squares with sizes from 1×1 to 5×5. How many of these squares contain the black center square?
Pick an answer.
AMC 10 2004 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 $5\times5$ grid holds squares of every size from $1\times1$ up to $5\times5$, all drawn along the grid lines. One cell — the black square in the exact middle — is marked. Count how many of these squares cover that middle cell.
Givens: The board is a $5\times5$ grid of unit cells.; Squares are drawn along grid lines in every size from $1\times1$ up to $5\times5$.; One cell — the black square in the exact center — is the target.; Answer choices: (A) $12$, (B) $15$, (C) $17$, (D) $19$, (E) $20$.
Unknowns: How many of the grid-aligned squares, across all sizes $1\times1$ through $5\times5$, contain the black center cell.
Understand
Restated: A $5\times5$ grid holds squares of every size from $1\times1$ up to $5\times5$, all drawn along the grid lines. One cell — the black square in the exact middle — is marked. Count how many of these squares cover that middle cell.
Givens: The board is a $5\times5$ grid of unit cells.; Squares are drawn along grid lines in every size from $1\times1$ up to $5\times5$.; One cell — the black square in the exact center — is the target.; Answer choices: (A) $12$, (B) $15$, (C) $17$, (D) $19$, (E) $20$.
Plan
Primary tool: #2 Make a Systematic List
Secondary: #1 Draw a Diagram, #7 Identify Subproblems, #5 Look for a Pattern
There are only five sizes of square, so the clean move is to split the count by size (Tool #7, Identify Subproblems) and tally each size separately in an organized list (Tool #2, Make a Systematic List) so nothing is missed or double-counted. A quick diagram (Tool #1) pins the black cell at the third column and third row, which stays fixed while we slide each square over it. For a given size, the trick is to see that 'covers the center' just means the square can start in only a few positions left-to-right and the same few up-to-down, so the placements form a small array. Counting those position-arrays for the five sizes produces the sequence $1,2,3,2,1$ of one-direction positions (Tool #5, Look for a Pattern), whose squares $1,4,9,4,1$ are the per-size counts to add.
Execute — Answer: D
2.G.A.2 Step 1 Locate the center square
- The black square is a single cell of the grid.
- Because the grid is $5$ by $5$, the middle cell sits in the third column from the left and the third row from the bottom.
- Every square we count — from $1\times1$ up to $5\times5$ — has to sit on top of this one cell.
- Marking that center cell keeps the target fixed while we slide different-sized squares over it.
💡 Fix the one cell everything has to cover before you start counting.
3.OA.A.1 Step 2 Turn 'covers the center' into a sliding count
- Think of a $k\times k$ square by where its left edge and bottom edge sit.
- Slide it left and right: it still covers the center column (column $3$) for only a few starting positions.
- Count those horizontal positions and call it $h$.
- By the exact same reasoning up and down, it covers the center row for the same number of positions, $h$.
- Since the left-right choice and the up-down choice are independent, the total number of placements is $h\times h = h^2$ — a square array of choices.
💡 Left-right choices times up-down choices counts every placement once, like cells in an array.
3.OA.A.1 Step 3 Count positions for each size
- For each size, count the horizontal positions $h$ that keep the square over column $3$.
- A $1\times1$ must sit exactly on the center: $1$ position.
- A $2\times2$ can start at column $2$ or $3$: $2$ positions.
- A $3\times3$ can start at column $1$, $2$, or $3$: $3$ positions.
- A $4\times4$ would like $3$ starts too, but starting at column $3$ would need columns $3,4,5,6$ — there is no column $6$ — so only columns $1$ and $2$ work: $2$ positions.
- A $5\times5$ fills the whole grid, so it has just $1$ position.
- That makes $h$ run $1,2,3,2,1$.
- Squaring each (horizontal $\times$ vertical) gives the placements $1,4,9,4,1$.
💡 The grid's edges choke off the biggest squares, so $4\times4$ and $5\times5$ get fewer spots than you'd first expect.
2.OA.A.1 Step 4 Add up all the sizes
- Every square that contains the black center is counted exactly once in these five groups, so add them up: $1 + 4 + 9 + 4 + 1 = 19$.
- That total counts all squares from $1\times1$ through $5\times5$ that cover the center cell, which matches choice (D).
💡 The five size-groups don't overlap, so a plain sum gives the grand total.
2.G.A.2 The black square is a single cell of the grid. Because the grid is $5$ by $5$, t 3.OA.A.1 Think of a $k\times k$ square by where its left edge and bottom edge sit. Slide 3.OA.A.1 For each size, count the horizontal positions $h$ that keep the square over colu 2.OA.A.1 Every square that contains the black center is counted exactly once in these fiv Review
Reasonableness: The counts $1,4,9,4,1$ are symmetric, which makes sense: a size-$k$ square and a size-$(6-k)$ square are hemmed in by the grid in mirror-image ways, so their position counts match ($1\leftrightarrow1$, $2\leftrightarrow2$). The total $19$ is also less than the grand total of all squares on the grid, $25+16+9+4+1 = 55$, as it must be, since only some squares cover the center. And $19$ is one of the listed choices, (D).
Alternative: Use complementary counting. Instead of counting squares that contain the center, count the ones that miss it and subtract from the total $55$. Only smaller squares can dodge the center: of the $25$ unit squares, $24$ miss it; of the $16$ two-by-two squares, $12$ miss it; and every $3\times3$, $4\times4$, and $5\times5$ square covers it, so those contribute $0$ misses. The misses total $24+12 = 36$, and $55 - 36 = 19$ — the same answer, (D).
CCSS standards used (min grade 3)
2.G.A.2Partition a rectangle into rows and columns of same-size squares and count to find the total number of them (Reading the $5\times5$ grid as rows and columns of unit squares and pinpointing the center cell at column 3, row 3.)3.OA.A.1Interpret products of whole numbers as the total number of objects in equal groups or arrays (Counting each square size's placements as (horizontal positions) $\times$ (vertical positions), a rectangular array of choices.)2.OA.A.1Use addition and subtraction within 100 to solve one- and two-step problems (Adding the five group counts $1 + 4 + 9 + 4 + 1$ to reach the grand total of $19$.)
⭐ To count how many squares cover one spot, slide each size left-right and up-down, multiply the two counts of positions that still cover the spot, then add the sizes up — and remember the grid's edges leave the biggest squares fewer places to sit.
⭐ To count how many squares cover one spot, slide each size left-right and up-down, multiply the two counts of positions that still cover the spot, then add the sizes up — and remember the grid's edges leave the biggest squares fewer places to sit.
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