Two numbers sit just below a hundred. Compute $97 \times 96$.

Use the base-100 method. The deficits from $100$ are $3$ and $4$. Cross-subtract either deficit from the other number, $97 - 4 = 93$ (or $96 - 3$), which gives the hundreds. The product of the deficits, $3 \times 4 = 12$, gives the last two digits. So $97 \times 96 = 9312$. It works because $(100 - a)(100 - b) = 100(100 - a - b) + ab$, and it is the fastest route for any pair of numbers just below a power of ten.

Takeaway: Near a power of ten: cross-subtract for the front, multiply the deficits for the back.