# SOLUTION: My first book required 1386 digits to consecutively number it pages. A.) How many pages does my book have? B.) How many times does the digit 6 appear in the book?

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 Question 349221: My first book required 1386 digits to consecutively number it pages. A.) How many pages does my book have? B.) How many times does the digit 6 appear in the book?Answer by Edwin McCravy(8999)   (Show Source): You can put this solution on YOUR website!```The number of counting numbers from counting integer A through counting integer B, inclusive of A and B is B-A+1 The 1 digit numbers are the integers between 1 and 9, inclusive of 1 and 9, is 9-1+1 = 9. The page numbers with 1 digit account for 9 digits The 2 digit numbers are the integers between 10 and 99 inclusive of 10 and 99, is 99-10+1 = 90. The page numbers with 2 digits account for 90*2 or 180 digits. So the page numbers from 1 through 99 account for 9 + 180 or 189 digits. The 3 digit numbers are the integers between 100 and 999 inclusive of 100 and 999, is 999-100+1 = 900. The page numbers with 3 digits would account for 900*3 or 2700. Since this is more than the 1386 digits in the page numbers of the book, we know that the remainder of the pages consist entirely of 3-digit numbers. Since 189 of the 1386 digits come from 1 and 2 digit numbers, the remaining 1386-189 or 1197 digits come from 3-digit numbers. Dividing 1197 by 3 gives 399, so the book contains 399 pages with 3-digit page numbers. Let N be the largest page number. Using the formula: The number of counting numbers from counting integer 100 through counting integer N, inclusive of 100 and N is N-100+1 or N-99. So this must equal 399. Setting these equal: N-99 = 399 N = 498 So the answer to A.) is 498 pages. -------------------------------------------- Now we need to find the number of 6's in the counting numbers through 498. First we get the number of 6's in all the one's digits: The page numbers with 6's in the last (ones) digits are 6,16,26,36,...,476,486,496 It's easy to see there are 50 of these by looking at the numbers obtained by removing the 6's from the right end, considering the beginning 6 to be 06. Removing the 6's from the right ends, we would have the numbers 0,1,2,3,...,47,48,49 So the one's digits of all the page number contain 50 6's. Next we look at the page numbers with 6's as their tens digits: They are 60-69,160-169,260-269,360-369,460-469 Each of these 5 groups consists of 10 numbers each. So the ten's digits of all the page number also contain 50 6's. None of the hundreds digits are 6's, so the total number of 6's is 50+50 or 100 6's. So the answer to B.) is 100 Edwin```