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.
To determine the last page number using 822 digits, we can break it down by the number of digits used for each range of page numbers. Pages 1 to 9 use 9 digits (1 digit each), pages 10 to 99 use 180 digits (90 pages × 2 digits), and pages 100 onward use 3 digits each. After using 189 digits for pages 1 to 99, there are 633 digits remaining for pages 100 and beyond, which accounts for 211 pages (633 ÷ 3). Thus, the last page is 99 + 211 = page 310.
1140
2890
5329 , i think
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.
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.
There are exactly 320 pages in 852 digits.
To determine the last page number using 822 digits, we can break it down by the number of digits used for each range of page numbers. Pages 1 to 9 use 9 digits (1 digit each), pages 10 to 99 use 180 digits (90 pages × 2 digits), and pages 100 onward use 3 digits each. After using 189 digits for pages 1 to 99, there are 633 digits remaining for pages 100 and beyond, which accounts for 211 pages (633 ÷ 3). Thus, the last page is 99 + 211 = page 310.
1140
There are 9 pages with a single digit (pages 1-9) = 9 digits There are 30 - 9 = 21 pages with two digits = 21 × 2 = 42 digits → There are 9 + 41 = 51 digits in total.
2890
315
An infinite number of pages are required to print an infinitely large prime number.
642
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.
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