Best Answer

There are exactly 320 pages in 852 digits.

Q: 852 digits used to number the pages of a book how many numbered pages does the book have?

Write your answer...

Submit

Still have questions?

Continue Learning about Algebra

There are 9 pages that use a single digit (pages 1-9), leaving 495 digits - 9 pages × 1 digit/page = 486 digits There are 90 pages that use 2 digits (pages 10-99), leaving 486 digits - 90 pages × 2 digits/page = 306 digits There are 900 pages that use 3 digits (pages 100-999); this would be 2,700 digits, so the number of pages is somewhere in the hundreds. 306 digits ÷ 3 digits/page = 102 pages in the hundreds. → total number of pages = 102 + 90 + 9 = 201 pages.

The pages of a book are numbered from 1 to 128. How many page numbers contain the digit 6?

No, they cannot. Even the concept of "all the digits of pi" is a problem. On the bright side, you don't have to get too many digits in before you have more than enough accuracy for any application you can imagine. Pi may be irrational, but it is reasonable nonetheless. The only digits I memorized to is 3.14159265358979323846264338327950288419716939937510.....that's all a 6th grade math book can give ya. The number being irrational means that it will go on and on forever. So, because of this is it not possible to know all the digits, because there will always be more to know.

1 1/2in thick and 500 pages

They are 44 and 45.

Related questions

642

This problem can be solved by applying the counting principle to digits in consecutive page numbers. To begin, we need to separate numbers into their numbers of digits in order to multiply the page numbers to find the number of digits needed to express them. Assuming that your book begins on page 1, there are 9 page numbers having one digit only (counting principle: 9 - 1 + 1 = 9). Since each of these pages is numbered with one digit, the number of digits used is 9 so far. Continue with the pages each numbered with two digits. These are pages 10-99, comprising 90 pages (99 - 10 + 1 = 90). Every page number multiplied by 2 digits each is 180. With the 9 digits coming from single-digit pages, the number of digits used so far is now 189. We can continue in the same manner for pages expressed with three digits, 100-999, but having 435 (624 total - 189 so far = 435) digits left, we probably won't be able to get through all the three-digit numbers. Also, a book is likely to have under a thousand pages. To find out how many pages are left, we divide the number of digits left by the number of digits needed for each page: 435/3 = 145 pages left. Since 145 pages only account for the pages numbered with three digits each, we need to add pages numbered with one and two digits each in order to find the total number of pages. Before 145 pages began to be accounted for, we had accounted for the digits of 99 pages (each numbered using one or two digits each), so the total number of pages is 99 + 145 = 244.

I'm going to go with 172. pages 1-9 = 9 digits pages 10-99 = 180 digiits leaves 219 digits each page from 100 on = 3 digits 219 /3 = 73 99 pages plus 73 page = 172

Let x be the number of pages in the book. Each page number has 3 digits, so the total number of digits used is 3x. We know that 3x = 492. Therefore, the number of pages in the book is x = 492 / 3 = 164 pages.

183 odd pages.

There are 9 pages that use a single digit (pages 1-9), leaving 495 digits - 9 pages × 1 digit/page = 486 digits There are 90 pages that use 2 digits (pages 10-99), leaving 486 digits - 90 pages × 2 digits/page = 306 digits There are 900 pages that use 3 digits (pages 100-999); this would be 2,700 digits, so the number of pages is somewhere in the hundreds. 306 digits ÷ 3 digits/page = 102 pages in the hundreds. → total number of pages = 102 + 90 + 9 = 201 pages.

183

315

9 digits used for pages 1 to 9 (9 pages * 1 digit)180 digits used for pages 10 to 99 (90 pages * 2 digits)453 digits used for pages 100 to 250 (151 pages * 3 digits)Total = 9 + 180 + 453 = 642 digits used.(Of course a trick answer is 10 as there are only 10 possible different digits in decimal).

1140

Pages 1 - 9 each have one digit, and so in total they use 9. Pages 10 - 99 each have two digits. There are 90 pages there, so we have now used 9 + (90 x 2) = 189 digits. Our book uses 264 - 189 = 75 digits more than this. That equates to 25 pages with three digits: pages 100 to 124 inclusive. So the book has 124 pages. Pages 1 - 9: 9 digits Pages 10 - 99: 180 digits. Pages 100 - 124: 75 digits. Total: 264 digits.

The 2989th digit of the final page in the book would be the 4 of page 1024.