Previous Problems of the Fortnight

We gratefully acknowledge the inspiration from books and websites on recreational mathematics (such as the Stanford Mathematics Problem Book) and mathematics problem contests.

Solution to the Problem of the Fortnight (14 October edition)

The problem:

Each page of an old book is numbered. The total number of digits in all the page numbers is 1890. How many pages does the book have?

(The Stanford Mathematics Problem Book, George Polya and Jeremy Kirkpatrick, Dover Publications)

A solution:

The book has 666 pages.

Pages 1 through 9 contribute one digit each, for a total of 9 digits.

Pages 10 through 99 contribute two digits each, for a total of 180 (2 times 90) digits, bringing the cumulative total of digits to 189.

Pages 100 through 999 contribute three digits each, for a total of 2700 (3 times 900) digits, which is too many, since our book only has a total number of 1890 digits in page numbers.

Subtracting 189 from 1890 gives us the number of digits contributed by the three-digit page numbers. So we have 1701 (=1890-189) digits for three-digit page numbers. Thus there are 567 (=1701/3) pages with three-digit page numbers. Add that to the number of pages with one-digit and two-digit page numbers, and we get 99+567 = 666 pages.

The winner:

Robert Penrod (first received, correct, complete solution)

Correct and complete solutions were also received from:

Jennifer Brustoski

Ryan Clarke

Rafe Donahue, ALUMNUS

Megan Giardina

Grant McCalister

Elizabeth Musco

Brady Powers

Scott Reighard

Thao Truong

Ben Wilson

Thank you for participating. Look for the new problem on Friday, October 28!