$10$
digit natural numbers do not contain the digit $6?$
Numbers cannot start with zero.

How many values can the first digit take?

For each of those values, how many different numbers that do not contain the digit
$6?$

Think of the different values each of the remaining digit can take.

The first digit can be 8 different options
$($
i.e. not$0$
and$6)$
and the rest can be anything but$6.$
The number of numbers without any 6s is$8 \times 9^9.$
(Note: We don’t expect you to calculate the final value!)