How can we partition the dots in a square grid into odd numbers of dots: ?

Question

Homi · Accepted Answer

Understand the Question

A square grid of dots (such as $1\times 1$ , $2\times 2$ , $3\times 3$ , etc.) can be built or partitioned in concentric L-shaped layers (also called gnomons).
Starting from a single dot in one corner ( $1$ ), each successive outer layer forms an "L" shape that contains the next odd number of dots ( $3, 5, 7, \dots$ ).
Thus, the total dots in an $n \times n$ square equal the sum of the first $n$ odd numbers: $1 + 3 + 5 + \dots + (2n - 1) = n^2$ .

Step 1 · Remove the Outermost Layer of a 3 × 3 Grid

Consider a $3 \times 3$ square grid consisting of $9$ dots in total. Diagram 1

Removing the outermost L-shaped layer of $5$ dots leaves a $2 \times 2$ square:

9 - 5 = 4

Step 2 · Remove the Next Layer

From the remaining $2 \times 2$ square containing $4$ dots, remove its outer L-shaped layer of $3$ dots. Diagram 2

4 - 3 = 1

Step 3 · Identify the Final Corner Dot

The remaining $1 \times 1$ square consists of just $1$ dot at the corner. Diagram 3

Thus, the $3 \times 3$ square grid is partitioned into consecutive odd numbers:

9 = 5 + 3 + 1

Step 4 · Generalize to Any Square Grid

This method applies to any square grid of size $n \times n$ by peeling off successive L-shaped borders.

For example, a $4 \times 4$ grid ( $16$ dots) is partitioned by removing layers of $7$ , $5$ , $3$ , and $1$ dots:

16 = 7 + 5 + 3 + 1

Answer

A square grid can be partitioned into odd numbers of dots ( $1, 3, 5, 7, \dots$ ) by dividing it into successive nested L-shaped layers starting from the single corner dot.

Common Mistakes

Splitting by rows or columns: Partitioning horizontally or vertically gives equal groups of $n$ dots instead of consecutive odd numbers ( $1, 3, 5, 7, \dots$ ).
Miscounting the corner dot in an L-shape: When counting an L-shaped layer of side length $k$ , ensure the corner dot is only counted once: each layer has $2k - 1$ dots.

Question 2

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