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

User Koraxis
by
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1 Answer

28 votes
28 votes

Answer:

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.

User Innat
by
3.1k points