For each natural number n, consider a square consisting of 2^(2n) smaller squares (i.e., a 2^n × 2^n square). If any one of the smaller squares is removed from the large square (not necessarily from the corner), then the resulting region can be completely covered by trominos (each made up of three small squares of the same size as the small squares in the region) in such a way that the trominos do not overlap.
Yes, for every natural number
✅ Key Concepts:
- A tromino is an L-shaped tile made up of 3 connected unit squares.
- We are working with a
square, which contains unit squares. - One square is removed, leaving
unit squares to be tiled.
🧠 The Proof (by induction):
Base case:
- A
square has 4 unit squares. - Removing one square leaves 3 squares — which perfectly matches the shape of a tromino.
- ✅ The base case holds.
Inductive step:
Assume the statement holds for some
Now consider a
- This square can be divided into four
quadrants. - The removed square lies in one of these four quadrants.
- By the inductive hypothesis, we can tile the affected quadrant after placing a tromino at the center of the full square that covers one square in each of the other three quadrants, near the center.
- Now each of the four quadrants has one square missing, so each can be tiled by the inductive hypothesis.
✅ This completes the inductive step.
📌 Conclusion:
By mathematical induction, any
Would you like a visual demonstration or code to simulate this tiling?