A printing company agreed to publish a book about Western history. When they were numbering the pages of the book, they noticed that they had used 2989 digits. How many pages long was the book they agreed to publish?
Can anyone help?
Why not look at some books you already have? Some of my books do not have numbers on some of the first pages, Title, dedication, publisher's note, first page and the numbering begins with page 4. Those pages without numbers are counted.
Other books begin with page 11, and some of the preceding pages are blank, or again have the title, dedication, etc.
Many of my reference books begin with Roman Numerals (I, II, III, IV, V, etc.) so you need to know if the word "digits" is including Roman Numerals.
I suspect it would be up to the Publishing Company exactly how a book is numbered! Also is that word counting page 29, for example, as 2 digits. That would mean that using 30 digits would mean only have 20 pages.
There were 1024 pages in the book, here is my reasoning
The number of single digit pages is 9, each one using 1 digit for a total of 9 digits each
the number of double digit pages is 90, (9x10), each page using 2 digits for a total number of digits of 180
the number of 3 digit pages is 900, (9x10x10), each page using 3 digits for a total of 2700 digits.
So at the end of page 999 we have 9+180+2700 or 2889 digits used.
So we still have 100 digits unaccounted for. The book must have run into the thousands in the page count, each of those pages using 4 digits.
So 100/4 = 25
So there must have been 25 pages starting with page 1000,
(from 1000 to 1024 is 25 pages)
1024
To find the number of pages in the book, we need to determine how many digits are there on each page.
Let's start by assuming that each page has the same number of digits.
Now, we know that there are 2989 digits in total. In order to determine the approximate number of pages, we need to find the maximum number of three-digit numbers (each number has three digits) that can be formed using the given 2989 digits.
Since each page has three digits, the number of three-digit numbers will be equal to the number of pages.
To find the maximum number of three-digit numbers, we divide the total number of digits (2989) by 3.
2989 divided by 3 equals 996 remainder 1.
This means that there are 996 three-digit numbers and one remaining digit.
Since the problem states that each page has the same number of digits, we assume that the remaining digit is on a separate page.
Therefore, the total number of pages in the book is 996 + 1 = 997.
So, the book they agreed to publish is 997 pages long.