AMC 8 · 2017 · #11

Easy mode Grade 4

Problem

Picture a square floor covered with square tiles. All the tiles are the same size, and they fit together perfectly like a grid.

Now picture both diagonals of the square drawn across the floor. The two diagonals cross all the tiles they pass through.

The total number of tiles that lie on either of the two diagonals adds up to $37$ .

How many tiles cover the whole floor?

Diagram

For small odd grids, count the tiles on both diagonals — the center tile is shared.

Pick an answer.

(A)

148

(B)

324

(C)

361

(D)

1296

(E)

1369

AMC 8 2017 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.

View mode:

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

Diagram

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 4)

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