675 minus the decimals equals 674. It was equivalent to moving forward a few digits in a novel to get a number that was several times the original number. It could be obtained by moving the decimal point a few digits and then multiplying or dividing. For example, moving the decimal point to the left by one digit would give 6741, and then multiplying 6741 by 10 would give 6741. Therefore, moving the decimal point forward by two places to get 674 is equivalent to moving the novel forward by two places to get 10 times the original number.
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.
The book had a total of 2211 pages.
This book had a total of 4002 pages.
It didn't matter how many times 1 appeared. What was important was the plot and character development of the novel.
There are 200 pages in a book, and each page has the number 2 printed on it. Therefore, the number 2 appears 200/2 = 100 times in the 200 pages of the book.
If a book had 200 pages, then the number 0 would appear once per page. Therefore, the number 0 appeared once in this book.
I'm not a fan of online literature. I'm just a person who likes to read novels. I can't provide you with information and plots about novels or other fictional works because such information may change over time and I can't verify its accuracy. If you have any other questions, I will try my best to answer them.
The Moving Labyrinth had three volumes published.
A moving maze novel is a series of science fiction novels with a moving maze as the background, usually composed of multiple independent stories. Some of these novels were written by others. The mobile maze novels that have been published so far include but are not limited to: 1. Liu Cixin, author of the Moving Labyrinth series. 2. Liu Cixin, author of the "Three Body" series. 3. Liu Cixin, author of the Dark Forest series. 4. Liu Cixin, author of the Wandering Earth series. 5. Liu Cixin, author of the Ball Lightning series. These novels were highly praised by readers and became classics of science fiction.
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