AMC 10 · 2016 · #15
Easy mode Grade 4The numbers 1,2,3,4,5,6,7,8,9 are placed in a 3×3 grid, one number per square. Any two numbers that come right after each other (like 4 and 5) must sit in squares that share a side. The four corner numbers add up to 18. What number is in the center square?
Pick an answer.
AMC 10 2016 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: The digits $1$ through $9$ are placed in a $3\times3$ grid, one per square, so that any two consecutive numbers sit in squares that share an edge. The four corner squares add up to $18$. Find the number in the center square.
Givens: Each of $1,2,3,4,5,6,7,8,9$ is used exactly once, one number per square; If two numbers are consecutive (like $4$ and $5$), their squares share an edge; The four corner numbers add up to $18$; Answer choices: (A) $5$, (B) $6$, (C) $7$, (D) $8$, (E) $9$
Unknowns: The number written in the center square
Understand
Restated: The digits $1$ through $9$ are placed in a $3\times3$ grid, one per square, so that any two consecutive numbers sit in squares that share an edge. The four corner squares add up to $18$. Find the number in the center square.
Givens: Each of $1,2,3,4,5,6,7,8,9$ is used exactly once, one number per square; If two numbers are consecutive (like $4$ and $5$), their squares share an edge; The four corner numbers add up to $18$; Answer choices: (A) $5$, (B) $6$, (C) $7$, (D) $8$, (E) $9$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #5 Look for a Pattern, #3 Eliminate Possibilities
Trying to actually fill the grid is slow and there are many arrangements. Instead, color the grid like a checkerboard (Tool #1). The chain $1\to2\to\cdots\to9$ only steps between edge-sharing squares, which always swap color, so the numbers' even/odd pattern lines up with the colors (Tool #5). Counting how many odd and even numbers there are then forces which color the corners-and-center must be (Tool #3) — and that pins the corner-plus-center total without ever building a full grid.
Execute — Answer: C
4.OA.C.5 Step 1 Color the grid like a checkerboard
- Shade the $3\times3$ grid in a checkerboard pattern, starting with a dark square in each corner.
- Then the four corners and the center are dark — five dark squares — and the four edge-middle squares are light.
- The key feature of a checkerboard: any two squares that share an edge have opposite colors.
💡 On a checkerboard every step to a neighbor flips the color, so corners and center share one color and the edges share the other.
2.OA.C.3 Step 2 Match parity to color
- Walk along the chain $1\to2\to3\to\cdots\to9$.
- Each step goes to an edge-sharing square, which flips the color, and each step also flips the number's parity (odd to even or even to odd).
- So color and parity flip in lockstep: every number sitting on a dark square has the same parity, and every number on a light square has the other parity.
💡 Two things that both flip at every single step stay perfectly matched the whole way.
2.NBT.B.5 Step 3 Force the odds onto the five dark squares
- Among $1$ through $9$ there are five odd numbers $\{1,3,5,7,9\}$ and four even numbers $\{2,4,6,8\}$.
- The dark squares (corners plus center) are five squares all of one parity.
- They cannot all be even — there are only four even numbers.
- So the dark squares must hold the five odd numbers.
- Adding them gives the corner-plus-center total.
💡 Five same-parity squares can't be filled by only four evens, so they must be the odds.
2.NBT.B.5 Step 4 Subtract to isolate the center
- The five dark squares are the four corners plus the center, and they total $25$.
- The four corners alone total $18$.
- The center is what remains, so subtract: $25 - 18 = 7$.
- So the center number is $7$, which is (C).
💡 Corners and center together are known, and the corners alone are known, so the leftover is the center.
4.OA.C.5 Shade the $3\times3$ grid in a checkerboard pattern, starting with a dark square 2.OA.C.3 Walk along the chain $1\to2\to3\to\cdots\to9$. Each step goes to an edge-sharing 2.NBT.B.5 Among $1$ through $9$ there are five odd numbers $\{1,3,5,7,9\}$ and four even n 2.NBT.B.5 The five dark squares are the four corners plus the center, and they total $25$. Review
Reasonableness: A concrete grid confirms it. The rows $\,3\,4\,5\,/\,2\,7\,6\,/\,1\,8\,9\,$ trace the chain $1\to2\to\cdots\to9$ along edge-sharing squares, so it is legal. Its corners are $3+5+1+9=18$ as required, and the center is $7$ — odd, exactly as the dark-square argument demands. The corner-plus-center cells $\{1,3,5,7,9\}$ sum to $25$, and $25-18=7$ matches.
Alternative: Test the choices using the parity fact. The center sits on a dark square, so it must be odd, immediately killing (B) $6$ and (D) $8$. Of the odd options, the corners-plus-center always sum to $1+3+5+7+9=25$, so the center equals $25-18=7$ no matter what; only (C) is consistent.
CCSS standards used (min grade 4)
4.OA.C.5Generate a number or shape pattern following a given rule (Building the checkerboard coloring as a shape pattern where neighboring squares always alternate color, separating the grid into five dark (corners + center) and four light (edges) squares.)2.OA.C.3Determine whether a group of objects has an odd or even number (Tracking that consecutive numbers alternate odd/even in lockstep with the alternating colors, so each color class holds numbers all of one parity.)2.NBT.B.5Fluently add and subtract within 100 (Adding the five odd numbers to get $1+3+5+7+9=25$, then subtracting the corner total to find the center $25-18=7$.)
⭐ Color the grid like a checkerboard: consecutive numbers must hop to a different color, so the five corner-and-center squares are exactly the five odd numbers — and $25-18$ leaves the center.
⭐ Color the grid like a checkerboard: consecutive numbers must hop to a different color, so the five corner-and-center squares are exactly the five odd numbers — and $25-18$ leaves the center.
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