Elementary Mathematics Dorofeev 〈PROVEN〉

Proof: Horizontal tromino covers cells (r,c), (r,c+1), (r,c+2). Their (row+col) mod 3 = (r+c) mod 3, (r+c+1) mod 3, (r+c+2) mod 3 → three consecutive integers mod 3 → all different residues 0,1,2. Same for vertical.

Now remove the top-left corner (1,1). Its color is (1+1) mod 3 = 2 mod 3 = Color 2? Wait — careful: (1+1)=2, so 2 mod 3 = 2 — yes, Color 2. So after removal: Color 0: 9 Color 1: 8 Color 2: 7 (since we removed one from Color 2) Each 1×3 tromino, no matter how you place it (horizontal or vertical), covers exactly one square of each color . elementary mathematics dorofeev

Thus, . 5. The Contradiction If 8 trominoes tile the shape, they would cover: 8 trominoes × 1 square of each color = 8 of Color 0, 8 of Color 1, 8 of Color 2. Now remove the top-left corner (1,1)

Try to visualize: the 5×5 board has 25 squares. Remove one corner → 24 squares. Each tromino covers 3 squares. 24 ÷ 3 = 8 trominoes needed. So numerically it’s possible. So after removal: Color 0: 9 Color 1:

Can you tile the remaining 24-unit shape with 1×3 "trominoes" (three squares in a straight line)?

But ? 2. The First Attempt You try. Place a tromino horizontally in the top row. Then another. You quickly get stuck — the missing corner leaves an awkward gap. After some attempts, you suspect it’s impossible .