2008 AMC 10B Problem 23

A $1$-foot border on all four sides leaves a painted rectangle of dimensions $(a - 2) \times (b - 2)$. The border is half the floor, so the painted part is the other half:
$$(a - 2)(b - 2) = \frac{ab}{2}.$$
Double and expand: $2ab - 4a - 4b + 8 = ab$, that is,
$$ab - 4a - 4b + 8 = 0.$$
Add $8$ to both sides to complete the factoring:
$$(a - 4)(b - 4) = 8.$$
With $b > a$, the factor pairs $(a-4, b-4)$ are $(1, 8)$ and $(2, 4)$, giving $(a, b) = (5, 12)$ and $(6, 8)$. Negative factor pairs such as $(-8, -1)$ or $(-4, -2)$ would make $a \le 0$, so they are excluded.

Check $(5, 12)$: painted $3 \times 10 = 30 = \tfrac12 \cdot 60$. Check $(6, 8)$: painted $4 \times 6 = 24 = \tfrac12 \cdot 48$.

The answer is $\boxed{\textbf{(B)}\ 2}$.

An equation of the form $xy + px + qy = r$ in integers is solved by adding the constant $pq$ so the left side factors as $(x + q)(y + p)$ (Simon's Favorite Factoring Trick); then the integer solutions come from the divisors of the right side. Geometric setups with borders or frames produce exactly this shape, so translate to algebra and factor rather than search.

• Using a-1 and b-1 for the painted rectangle (a 1-foot border removes 1 foot from each side, so 2 from each dimension).
• Counting the factorization $(4, 2)$ as a second solution despite the requirement $b > a$, or forgetting to check that negative factor pairs give invalid dimensions.