The Garden Path Tile Pattern
A student is arranging square tiles to border a rectangular garden. The border is one tile wide. For a garden that is $m$ tiles wide and $n$ tiles long (measured in tile lengths), the number of border tiles is $2m + 2n + 4$. The student draws gardens with $m = 3$ and varying $n$: for $n = 1, 2, 3$ the border counts are 12, 14, 16. The student claims the border count always increases by 2 when $n$ increases by 1, and that a border of exactly 40 tiles is possible for some whole-number $m$ and $n$. A classmate says a border of 41 tiles is also possible.
1. Verify the pattern for $m = 3$ and explain why the border count increases by 2 each time $n$ increases by 1.
Read the worked answer
Formula: $2m + 2n + 4$ with $m = 3$ gives $6 + 2n + 4 = 2n + 10$.
- $n = 1$: $2 + 10 = 12$
- $n = 2$: $4 + 10 = 14$
- $n = 3$: $6 + 10 = 16$
When $n$ increases by 1, the expression $2n + 10$ increases by $2(1) = 2$. This matches the drawing because each extra row of garden adds one tile on each of the two long sides, i.e. 2 tiles, while the four corner tiles stay fixed.
The pattern is correct.
2. Determine whether a border of exactly 40 tiles is possible with whole-number $m \ge 1$ and $n \ge 1$. Find all such pairs.
Read the worked answer
Set $2m + 2n + 4 = 40$.
$$2m + 2n = 36 \implies m + n = 18$$
With $m \ge 1$ and $n \ge 1$, the whole-number pairs are:
$(1,17), (2,16), (3,15), \ldots, (17,1)$ - 17 pairs in total.
Yes, 40 tiles is possible, in 17 different ways.
3. The classmate claims a border of 41 tiles is possible. Decide whether this is true, and explain using parity (odd/even reasoning).
Read the worked answer
Set $2m + 2n + 4 = 41$.
$$2m + 2n = 37$$
The left side $2m + 2n = 2(m+n)$ is always even, but 37 is odd. An even number cannot equal an odd number, so there are no whole-number solutions.
Equivalently, $2m + 2n + 4$ is always even (even + even + even), so the border count can never be odd.
The classmate is wrong: 41 tiles is impossible.