## 5. Greater than 6

How many three-digit natural numbers have at least one digit greater than 6?

Result: $\quad 606$
## Solution, first method

There are a total of 900 three-digit numbers. Let's calculate how many of them do not have at least one digit greater than 6, i.e., those for which each digit is less than or equal to 6. For such numbers, the hundreds digit can be chosen in 6 ways (the hundreds digit can be $1,2,3,4,5,6$), and the tens and units digits can be chosen in 7 ways (the tens and units digits can be $0,1,2,3,4,5,6$), so there are $6 \cdot 7 \cdot 7=294$ such numbers.

Therefore, the number of desired numbers is 900 - $294=606$.

## Solution, second method