Multiplying Two Numbers Whose Units Add to 10 and the Rest is the Same

This is a generalization of the 'Squares Ending in 5' trick. For two numbers where the units digits add to 10 and the remaining leading digits are the same (e.g., 68 and 62), there's a shortcut.

Let the numbers be $10a+b$ and $10a+c$ where $b+c=10$. The product is $100a(a+1) + bc$.

Procedure:

• The last two digits of the answer are the product of the units digits ($b \times c$).
• The preceding digits are the leading digit(s) ($a$) multiplied by one greater than itself ($a+1$).

For example, for $68 \times 62$: The last two digits are $8 \times 2 = 16$. The preceding digits are $6 \times (6+1) = 42$. The answer is 4216.