AMC 10 · 2016 · #12

Grade 4 geometry-2d
parityparity-coloringlogical-deduction parity-coloring ↑ Prerequisites: parity
📏 Medium solution 💡 3 insights
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Problem
Consecutive numbers must sit in touching squares and the corners have a known total. Find the centre entry.

Pick an answer.

(A)
5
(B)
6
(C)
7
(D)
8
(E)
9
How to solve
Strategy Draw a Diagram

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→2→…→9 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.

1STEP 1

Color the grid like a checkerboard

A checkerboard colouring splits the grid five to four.

● & ° & ● ; ° & ● & ° ; ● & ° & ● ; ●=dark (5), °=light (4)
2STEP 2

Match parity to color

Consecutive numbers must change colour.

1_●→2_°→3_●→4_°→…
3STEP 3

Force the odds onto the five dark squares

So all the odd numbers sit on one colour.

1+3+5+7+9 = 25
4STEP 4

Subtract to isolate the center

Subtracting the corners gives 7, choice (C).

center = 25 - 18 = 7 = (C)
Answer
7
A concrete grid confirms it. The rows 3 4 5 / 2 7 6 / 1 8 9 trace the chain 1→2→…→9 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.
💡Key takeaway

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
  • Match parity to color
  • Force the odds onto the five dark squares
  • Subtract to isolate the center