Question

If it takes 909 digits to number the pages of a book starting with page 1, how many pages are in the book?

Answer 1

There would be
`339` pages in this book.

Explanation:

Important: there are
`(x - y + 1)` pages between page
`y` and page
`x` (
`x > y`.)

There are
`(9 - 1 + 1) = 9` pages between page
`\verb!1!` and
`\verb!9!`. Each page takes
`1\!` digit to number. That would be
`9 * 1 = 9` digits.

There are
`(99 - 10 + 1) = 89 + 1 = 90` pages between page
`\verb!10!` and
`\verb!99!`. Each page takes
`2` digits to number. That would be
`90 * 2 = 180` digits.

So far,
`9 + 180 = 189` digits are used. There are
`909 - 189 = 720` more digits to consider.

Each page between
`\verb!100!` and
`\verb!999!` takes
`3` digits to number. Hence,
`720` digits would number
`720 / 3 = 240` pages in that range.

Important: let
`x` denote the number of the last page in this book.
`(x - 100 + 1)` would denote the number of pages that take
`3` digits each to number.

Reasoning above shows there are
`240` of these pages in total. That is:

`(x - 100 + 1) = 240`.

Hence:

`x = 240 + 99`.

`x = 339`.

In other words, the last page of this book is numbered
`\verb!339!`. There would be
`339` pages in this book.