2004 AMC 10B Problem 2

"At least one $7$" is easiest to handle through its opposite, "no $7$ anywhere."

There are $90$ two-digit numbers ($10$ through $99$). To build one that avoids the digit $7$: the tens digit can be any of $1$–$9$ except $7$, which is $8$ options, and the units digit can be any of $0$–$9$ except $7$, which is $9$ options. That gives $8 \cdot 9 = 72$ numbers with no $7$.

Everything else contains a $7$:
$$90 - 72 = 18.$$

The answer is $\boxed{\textbf{(B)}\ 18}$.

"At least one" conditions are usually cleaner from the complement, because "none" is a simple product of independent digit choices, while counting directly forces you to manage an overlap. The overlap here is the number $77$, which belongs to both the "tens digit is $7$" group and the "units digit is $7$" group.

Direct count with overlap removed: the numbers $70$–$79$ contribute $10$; the numbers $17, 27, \dots, 97$ contribute $9$; the number $77$ appears in both lists, so the total is $10 + 9 - 1 = 18$.

• Adding 10 numbers with tens digit 7 to 9 numbers with units digit 7 and answering 19, counting 77 twice.
• Allowing $0$ as a tens digit (or forbidding it as a units digit) in the complement count, which changes $8 \cdot 9$ to $9 \cdot 9$ or $8 \cdot 8$.