It takes 2202 digits to arrange the page number of a novel. How many pages does this book have?

Answer 1

Assuming that the book has $n$pages, then the page number needs to contain $n$numbers, each number representing a page number.

Since the page number needs to be numbered, the total number of digits of the page number must be the power of $2$, which is $2k $, where $k$represents the number of digits of the page number. According to the question, the page number needs to use $2202$numbers, so the value of $k$should be a factor of $2202$, which means that $k$can be $1247142872144288 $.

For any $k$, you can use the Enumeration Method to calculate how many pages you need to use $2k $numbers. For example, when $k=1$, there is $n=2202= 2 ^4× 72$, so the book has a total of $2202+72=2274$pages. When $k=2$, there is $n= 2 ^7 times 144$, so this book has a total of $2 ^7 times 144+144=25108$pages. When $k=4$, there is $n= 2 ^14 times 288$, so this book has a total of $2 ^14 times 288+288=32064$pages. When $k=7$, there is $n= 2 ^28×72 $, so this book has a total of $2 ^28×72 +72=35904$pages. When $k=14$, there is $n= 2 ^44 times 288$, so this book has a total of $2 ^44 times 288+288=46608$pages. When $k=28$, there is $n= 2 ^72> times 144$, so the book has a total of $2 ^72> times 144+144=29472$pages. When $k=72$, there is $n= 2 ^144 times 288$, so this book has a total of $2 ^144 times 288+288=331728$pages.

Therefore, it can be concluded that this book has a total of $2274+32064+46608+29472+33172+46832+35904+46608+29472+35904+25108+2202=298768$pages.