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Q: How many digits used to number 250 page book?

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381 digits.

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.

Page 43. This depends on the copy of your book, too.

If all pages are numbered (usually page 1 is not numbered) then 537.

Set up equation; x = first page number & x+1 = second page number. So, x + x + 1 = 369 or x = 184. X+1 = 185 which is the larger page number.

Related questions

That's going to depend on the subject matter of the book. Besides the digits used to number the pages, an arithmetic book will have a lot more digits in it than, for example, a novel or a book of poetry has.

2

Three digits.

381 digits.

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

There are 64 pages (100 - 163) that need 3 digits per page; 64 x 3 = 192. There are 100 pages (10 - 99) that need 2 digits per page; 100 x 2 = 200.There are 9 pages (1 - 9 ) that need one digit per page; = 9.So 9 + 200 + 192 = 401 digits.[Assuming that the last page, an even-numbered page, has no number.]

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.

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.

Assessor Parcel Numbers (APN) are 8 digits long. The standard formatting for the numbers is 000-000-00. The book number is the first three digits. The page number is the next two digits. Your APN can be found on your value notice or tax bill.

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

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.

200