Alexander, from Wilson's School, wrote:

Phoebe first drew one “down,” and then as many “inverted Cs” as needed to get seven squares. Therefore:

- When drawing only seven squares you draw one “down” and seven “inverted Cs.” There were 22 matchsticks.
- When drawing 25 squares, there would still be one “down,” but 25 “inverted Cs.” The total number of matchsticks would be 76.
- When drawing 100 squares, again there would be only one “down,” but 100 “inverted Cs.” The total number of matchsticks would be 301.
- When drawing 1,000,001 squares, there would be 1 “down” and 1,000,001 “inverted Cs.” There would be 3,000,004 matchsticks.
- When drawing n squares, there would be one “down” and n “inverted Cs.” There would be 3n + 1 matchstick(s). This is because there is one matchstick at the beginning that forms the “down,” and then each of the “inverted Cs” takes up three matchsticks, so you multiply the number of inverted Cs by three and then add 1.

Alice first drew seven “alongs” at the top, and then seven “alongs” at the bottom. She then used eight “downs” to connect the gaps in between two parallel “alongs.” Therefore:

- For seven squares you draw fourteen “alongs” and eight “downs.” The total number of matchsticks is 22.
- When drawing 25 squares, you draw 50 “alongs” and 26 “downs,” making the total number of matchsticks 76.
- When drawing 100 squares, you draw 200 “alongs” and 101 “downs,” making the total number of matchsticks 301.
- When drawing 1,000,001 squares, you draw 2,000,002 “alongs” and 1,000,002 “downs,” making the total 3,000,004 matchsticks.
- When drawing n squares, you draw 2n “alongs” and n + 1 “downs,” making the total 3n + 1 matchsticks. This works because for each square, there are two “alongs,” and like an “inverted C,” you have one “down” for each square, but then you add one more “down,” because like Phoebe’s method, you need one extra line at the end. Adding the two formulas together, 2n + n + 1 = 3n + 1, which is the same as Phoebe’s formula.

Luke first drew one “square,” and then six “inverted Cs” to make seven squares. Therefore:

- You need one “square” and six “inverted Cs” to make seven squares. This gives you a total of 22 matchsticks.
- Similarly, with 25 squares, you draw one “square,” but 24 “inverted Cs,” which gives you a total of 76 matchsticks.
- With 100 squares, you still draw one “square,” but 99 “inverted Cs,” which gives you a total of 301 matchsticks.
- With 1,000,001 matchsticks, again you draw one “square,” but you draw 1,000,000 “inverted Cs,” giving you a total of 3,000,004 matchsticks.
- With n matchsticks, you draw one “square,” and n – 1 “inverted Cs,” giving you a total of 3n + 1 matchsticks:

4 + 3(n – 1) = 4 + 3n – 3 = 3n + 1,

which is the same as all of the other methods.

Great! Laeticia, from Woodbridge High School, also gave correct general formulas here. She went on to comment on one of our later puzzles:

Growing rectangles:

If the height is h and the width is w, then the perimeter is 2h + 2w. Then there are (h+1)(w+1) dots, and w(h+1) + h(w+1) lines.

Niharika solved the rest of our questions:

**L-shapes:**

Each L-shape has an 'inner' L-shaped line and an 'outer' L-shaped line. If an L-shape has height and width n, then the outer L has length 2n and the inner L has length 2(n-1). There are 2 lines remaining on the ends of the L, so altogether the perimeter is 4n.

There are 2(n-1) lines left inside the L-shape, so the shape is made of 6n - 2 lines.

The number of squares in an L-shape is 2n-1: think of each L as two rectangles of height 1 and width n that overlap in one square.

**Two squares:**

Think of these as two separate overlapping squares, each with $n^2$ dots. They overlap in one dot, so in total there are $2n^2 - 1$ dots, and so $2n^2 - n - 1$ white dots.

In each square there are 2n(n-1) lines, and the lines never overlap, so in total there are 4n(n-1) lines.

**Square of squares:**

We can split a pattern like this of side length n up into four rectangles of height 1 and length n-1, so there are 4(n-1) edge squares.

There are 4n lines making up the outer square and 4(n-2) lines making up the inner square. There are 4(n-1) lines left in the middle. This gives 12(n-1) lines in total.

**Dots and more dots:**

Inside the squares there are $n^2$ dots, and on the vertices there are $(n+1)^2$ dots, so in total there are $2n^2 + 2n + 1$ dots.

In each row there are n lines, and there are n+1 rows, so the rows contribute n(n+1) lines. Similarly the columns contribute n(n+1) lines, so together there are 2n(n+1) lines.

**Rectangle of dots:**

There are 4n horizontal lines and 3n vertical lines, so 7n lines in total.

On each row there are 2n+1 dots, and there are n+1 rows, so there are (2n+1)(n+1) dots.

Fantastic! Thank you all.