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?
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
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) $1369$
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.5Step 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.
💡 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.5Try smaller floors first. Draw a $2 \times 2$, $3 \times 3$, $4 \times 4$, and $
[2]
#5 4.OA.C.5Look at the four counts $4, 5, 8, 9$ side by side. The even-$n$ rows give $2n$ (
[3]
#5 2.OA.C.3The problem gives $37$ diagonal tiles. Since $37$ is odd, $n$ must be odd, so we
[4]
#9 4.NBT.B.5Compute 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!
⭐ This AMC 8 problem only needs Grade 4 pattern-finding and two-digit multiplication you already know!
