## 3. Even and Odd

How many three-digit natural numbers have at least one even digit and at least one odd digit? Zero is an even digit.

## Result: $\quad 675$
## First solution.

There are $9 \cdot 10 \cdot 10=900$ three-digit numbers. Let $p$ be an even digit, and $n$ be an odd digit. Then $p \in\{0,2,4,6,8\}$ and $n \in\{1,3,5,7,9\}$.

Three-digit numbers where all digits are even have $4 \cdot 5 \cdot 5=100$ because the hundreds digit is different from zero. Three-digit numbers where all digits are odd have $5 \cdot 5 \cdot 5=125$.

If we exclude from the set of all three-digit numbers those that have all even or all odd digits, the remaining numbers will have at least one even and at least one odd digit.

Their number is $900-100-125=675$.

There are 675 three-digit numbers where at least one digit is even and at least one digit is odd.