How many integers exist from 0 to 999999, in the decimal representation of which there are no two adjacent identical digits?

#
The number of $n$-digit numbers with such a property is $9^{n}$: the first place can be any of 9 digits (all except 0), the second place can also be any of 9 digits (all except the first one), and so on.

## Answer

$10+9^{2}+9^{3}+9^{4}+9^{5}+9^{6}=\frac{9^{7}-1}{8}=597871$ numbers.