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381 digits.

Q: How many digits are used for printing a page numbers of a book that has 163 pages?

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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.

299 pages.

1140

This depends on exactly what you mean by "100 pages" If you mean exactly 100 enumerated page sides, then you can say: The first nine pages would have 1 digit, so that's 9. the next 90 pages would have two digits, so that's 180. The last page would have three digits. Add them all together and we get: 9 + 180 + 3 = 192 If on the other hand mean 100 pieces of paper (200 page sides), then we have: The first nine pages would have 1 digit, so that's 9. the next 90 pages would have two digits, so that's 180. The next 101 pages would have 3 digits, so that's 303. Adding those we get: 9 + 180 + 303 = 492

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Related questions

Three digits.

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.]

First 9 pages = 9 digit. That leaves 142 digits. @ 2 digit per page, that is 142/2 = 71 pages with 2-digit numbers. So, 9 pages with 1-digit numbers + 71 pages with 2-digit numbers = 80 pages.

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.

There are exactly 320 pages in 852 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.

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

299 pages.

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.

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).

315

453 0-9 9 digits 10-99 180 digits 100-250 453 digits