2004 AMC 10B Problem 2

"At least one $7$" could mean a $7$ in the tens place, in the ones place, or in both. Rather than juggle those, count the opposite: numbers with no $7$ at all.

1. There are $90$ two-digit numbers, from $10$ to $99$.
2. Build a two-digit number with no $7$. The tens digit can be $1$ to $9$ but not $7$: that is $8$ choices. The ones digit can be $0$ to $9$ but not $7$: that is $9$ choices.
3. Every tens choice can go with every ones choice, so there are $8 \times 9 = 72$ numbers with no $7$.
4. All the other two-digit numbers have at least one $7$:
$$90 - 72 = 18.$$

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

When a question says "at least one," the opposite ("none") is often a simple multiplication, because each digit is chosen on its own. Counting directly means you have to watch for numbers that fit two ways at once, here the number $77$.

Count directly, but watch for the overlap.

• The numbers $70, 71, \ldots, 79$ have a $7$ in the tens place: $10$ numbers.
• The numbers $17, 27, \ldots, 97$ have a $7$ in the ones place: $9$ numbers.
• The number $77$ is on both lists, so it was counted twice. Take one copy away: $10 + 9 - 1 = 18$.

"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 $\boxed{\textbf{(B)}\ 18}$.