AMC 8 · 2017 · #11

Grade 4 geometry-2dpattern

Archetype Lattice / Grid Pattern from Small Cases

perfect-squaresparitypattern-recognitionspatial-visualization pattern-recognitioneasier-related-problem ↑ Prerequisites: multi-digit-arithmeticparity

📏 Medium solution 💡 3 insights

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Problem

A square-shaped floor is covered with congruent square tiles. If the total number of tiles that lie on the two diagonals is 37, how many tiles cover the floor?

$\textbf{(A) }148\qquad\textbf{(B) }324\qquad\textbf{(C) }361\qquad\textbf{(D) }1296\qquad\textbf{(E) }1369$

Pick an answer.

(A)

148

(B)

324

(C)

361

(D)

1296

(E)

1369

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Toolkit + CCSS Solution

Understand

Restated: A square floor is paved with congruent square tiles, forming an $n \times n$ grid. The tiles sitting on the two main diagonals of the floor total $37$. Find the total number of tiles covering the whole floor.

Givens: The floor is a square grid of $n \times n$ congruent square tiles; The total number of tiles lying on the two main diagonals is $37$; Answer choices: (A) $148$, (B) $324$, (C) $361$, (D) $1296$, (E) $1369$

Unknowns: The total number of tiles on the floor, which equals $n^2$

Understand

Plan

Primary tool: #9 Solve an Easier Related Problem

Secondary: #5 Look for a Pattern, #1 Draw a Diagram

We do not know the side length $n$, and trying to attack $n^2$ directly with the value $37$ is awkward. Tool #9 (Easier Related Problem) says: shrink the floor first — draw a $2 \times 2$, $3 \times 3$, $4 \times 4$, $5 \times 5$ grid (Tool #1) and count the diagonal tiles in each. Tool #5 (Look for a Pattern) then reveals the rule: odd $n$ gives $2n - 1$ diagonal tiles, even $n$ gives $2n$. Because $37$ is odd, $n$ must be odd, and $2n - 1 = 37$ pinpoints $n$. We finish by computing $n^2$.

Execute — Answer: C

#9 Solve an Easier Related Problem 4.OA.C.5 Step 1

Try smaller floors first.
Draw a $2 \times 2$, $3 \times 3$, $4 \times 4$, and $5 \times 5$ grid and count the tiles that lie on either of the two diagonals.
For even sides, the diagonals cross between tiles, so no tile is shared; for odd sides, both diagonals run through the single center tile.

$$n=2:\;2+2=4 \quad n=3:\;3+3-1=5 \quad n=4:\;4+4=8 \quad n=5:\;5+5-1=9$$

💡 Generating the first few cases from a rule ("count tiles on the diagonals") is exactly what Grade 4 pattern-generating practice asks for.

#5 Look for a Pattern 4.OA.C.5 Step 2

Look at the four counts $4, 5, 8, 9$ side by side.
The even-$n$ rows give $2n$ (always even); the odd-$n$ rows give $2n - 1$ (always odd).
So the parity of the total diagonal count tells you the parity of $n$.

$$\text{diagonal tiles} = \begin{cases} 2n & n \text{ even} \\ 2n - 1 & n \text{ odd} \end{cases}$$

💡 Spotting that the totals split into an "even family" and an "odd family" is a Grade 4 pattern observation, no algebra needed.

#5 Look for a Pattern 2.OA.C.3 Step 3

The problem gives $37$ diagonal tiles.
Since $37$ is odd, $n$ must be odd, so we use the formula $2n - 1 = 37$.
Solve by adding $1$ to both sides and then halving.

$$2n - 1 = 37 \;\Rightarrow\; 2n = 38 \;\Rightarrow\; n = 19$$

💡 Recognizing $37$ as odd is a Grade 2 odd/even check; then applying the pattern formula (Tool #5) backwards gives $n = 19$ in one short step.

#9 Solve an Easier Related Problem 4.NBT.B.5 Step 4

Compute the total number of tiles, which is the area of the $19 \times 19$ grid.
Use the standard two-digit by two-digit multiplication: $19 \times 19 = (20 - 1)(20 - 1) = 400 - 20 - 20 + 1 = 361$.

$$n^2 = 19 \times 19 = 361 \;\Rightarrow\; \textbf{(C)}$$

💡 Multiplying two two-digit numbers using place-value strategies (here, $(20-1)^2$) is exactly the Grade 4 multi-digit multiplication standard.

[1] #9 4.OA.C.5 Try smaller floors first. Draw a $2 \times 2$, $3 \times 3$, $4 \times 4$, and $

[2] #5 4.OA.C.5 Look at the four counts $4, 5, 8, 9$ side by side. The even-$n$ rows give $2n$ (

[3] #5 2.OA.C.3 The problem gives $37$ diagonal tiles. Since $37$ is odd, $n$ must be odd, so we

[4] #9 4.NBT.B.5 Compute the total number of tiles, which is the area of the $19 \times 19$ grid.

Review

Reasonableness: Sanity check the answer choices first. (A) $148$ is not a perfect square, so it cannot be $n^2$ for an integer $n$ — eliminate. (D) $1296 = 36^2$: with $n = 36$ (even) the diagonals would give $2 \times 36 = 72$ tiles, not $37$ — eliminate. (E) $1369 = 37^2$: with $n = 37$ (odd) the diagonals would give $2 \times 37 - 1 = 73$ tiles, not $37$ — eliminate. (B) $324 = 18^2$: $n = 18$ is even, giving $36$ diagonal tiles, not $37$ — eliminate. (C) $361 = 19^2$: $n = 19$ is odd, giving $2 \times 19 - 1 = 37$ diagonal tiles. Matches.

Alternative: Tool #3 (Eliminate Possibilities) on the choices directly: for each choice $C$, set $n = \sqrt{C}$ and check whether $2n$ or $2n - 1$ equals $37$. Only $C = 361$ (with $n = 19$, odd, giving $2 \times 19 - 1 = 37$) survives. This back-substitution approach is faster than deriving the formula if you spot that all five choices are nearly perfect squares.

CCSS standards used (min grade 4)

4.OA.C.5 Generate a number or shape pattern following a given rule (Producing the small-case counts $4, 5, 8, 9$ from the rule "count tiles on the two diagonals of an $n \times n$ grid" and noticing the parity-based pattern.)
2.OA.C.3 Determine whether a group of objects has an odd or even number (Observing that $37$ is odd, which forces $n$ to be odd and selects the formula $2n - 1$ over $2n$.)
4.NBT.B.5 Multiply a whole number of up to four digits by a one-digit whole number, and multiply two two-digit numbers (Computing the final answer $19 \times 19 = 361$ using a place-value strategy such as $(20-1)^2$.)

⭐ This AMC 8 problem only needs Grade 4 pattern-finding and two-digit multiplication you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 4)

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