. How many zeros are there in the number:

$12345678910111213141516171819202122 \ldots 20062007$
. Among the single-digit numbers (i.e., those from 1 to 9), there are no zeros. Among the two-digit numbers, there are nine zeros: one at the end of 10, one at the end of 20, and so on up to 90. Now let's see what happens with three-digit numbers. A zero can obviously only appear in the second or third position. There is a zero in the second position in all numbers of the form $x 0 y$ where $x$ is a digit between 1 and 9, and $y$ is a digit between 0 and 9. This makes 90 such numbers. We reason similarly for the third zeros and find 90 as well. The same reasoning allows us to count the zeros that appear in the numbers from 1000 to 1999: those that appear in the second (resp. third, resp. fourth) position are in the number of $1 \times 10 \times 10=100$. This makes a total of 300. There remain the zeros that appear in the numbers between 2000 and 2007, which can be counted by hand; we find 17. In the end, the number sought is $9+90+90+300+17=506$.