Pages in a book are numbered in typical fashion, starting with !. The folio for page 10 will contain the tenth and eleventh digits necessary to paginate the book. On what page will the 2009th digit occur?

To find the page on which the 2009th digit occurs, we need to first determine the pattern of how the digits are numbered.

In this case, we are told that the first page is numbered as 1 and the folio for page 10 contains the tenth and eleventh digits. From this information, we can infer that each page consists of two digits.

To find the number of digits on each page, we will subtract the previous page's total number of digits from the current page's total number of digits. On page 1, there is only one digit (since it's the first page). On page 2, there are two digits (since we have already used one digit). Similarly, on page 3, there are two digits. This pattern repeats, with every even-numbered page containing two digits and every odd-numbered page containing one digit.

Now, we need to determine how many pages it takes to reach the 2009th digit. To do this, we can create a sum of the digits on each page until we reach or exceed 2009.

Let's calculate:

Page 1: 1 digit (cumulative total = 1)
Page 2: 2 digits (cumulative total = 3)
Page 3: 2 digits (cumulative total = 5)
Page 4: 2 digits (cumulative total = 7)

Continuing this pattern:

Page 5: 1 digit (cumulative total = 8)
Page 6: 2 digits (cumulative total = 10)
Page 7: 2 digits (cumulative total = 12)
Page 8: 2 digits (cumulative total = 14)

Based on this pattern, we can see that it takes 4 pages to reach the first 8 digits and their cumulative total is 14. We need to find the page where the cumulative total reaches 2009.

Since we have 14 digits after 4 pages, we need to go through approximately 144 pages (14 * 10) to get 140 digits. This will put us at a cumulative total of 2000 digits.

The 2009th digit is the 9th digit on the next page, which is 145 (since we added 144 pages).

Therefore, the 2009th digit occurs on page 145.