There were 100 pages in the main body of a novel. How many words would it take to compile these pages?

It was about 6060 pages. <strong></strong> If you like short stories, you can read "I'm Urged for More Today" and "Hey, How Are You?" If you like fantasy novels, you can read "My Magic Weapon Life and Death Book" and "Emperor Wu Ji." If you like games and virtual online games, you can read "Three Kingdoms Mercenary Corps." If you like modern romance and aristocratic families, you can read "Cute Wife: President, Doting on Her" and "Mr. President, 7-Page Rules." If you liked historical fiction, you could read "The Unscrupulous Emperor". If you liked light fiction, you could read "Dragon: My System Ran On The First Day." I hope you like this fairy's recommendation. Muah ~😗

20 pages. Just divide the total number of words (7000) by the number of words per page (350), so 7000 / 350 = 20.

Four pages. 1000 divided by 250 is 4.

This novel had about 238 pages x 625 words per page = 13240000 words. Therefore, this novel had about 130,000 words.

The number of words in the novel is 750. We know that each page contains 250 words. By using basic arithmetic, we divide the total number of words in the novel by the number of words per page. So, 750 / 250 = 3. Thus, the 750 - word novel would be 3 pages.

10,000 words usually equaled 100000 pages. A page usually contains 250 words, so a book with a million words will have about 2500 pages. However, this is only a rough estimate because the number of pages and words in a book may vary according to the author, the publishing house, the time of publication, the subject matter, and other factors. In addition, the number of words and pages in a book may also vary due to factors such as the length of the text and the compression method.

The number of MW pages equivalent to 1 novel page can vary widely. It could be anywhere from half a MW page to two or more, depending on formatting and content density.

It really depends on a few factors like font size, line spacing, and margins. But a rough estimate could be around 75,000 to 150,000 words, which might translate to 300 to 600 typed pages.

Well, that really depends. Different publishers might have different page counts for the same novel. It could range anywhere from a couple hundred to several hundred pages.

This problem could be solved through mathematical methods. Assuming that the novel has $n$pages, then each page has $p$numbers, where $1'le p 'le n$. According to the page number of the question, a total of $297$numbers can be listed as follows: $$n\times p + n - 1 = 297$$ To simplify it: $$n(p+1) = 297 - 1 = 296$$ Since $n$is an integral,$p+1$must be a multiple of $296$. At the same time, since $1'le p 'le n$,$p+1$must be a multiple of $12' ldotsn'$. Therefore, the following restrictions can be obtained: $$p+1> text {is a multiple of $1$but not a multiple of $2$} p+1> text {is a multiple of $2$but not a multiple of $3$}& ldots p+1> text {is a multiple of $n$but not a multiple of $n-1 $}$$ According to these constraints, the value range of $p+1$can be obtained: $$135791113\ldotsn$$ Substituting these values into the equation $n(p+1) = 297 - 1$gives: $$n(n+1) = 297 \times (n+1)$$ To simplify it: $$n^2 + n - 296 = 0$$ By solving this second order equation, one could get: $$n = \frac{296\pm\sqrt{296^2-4\times1\times296}}{2\times1} = \frac{296\pm294}{2}$$ Since $n$is an integral number,$n$can only take two values: $$n = 44 n = 43$$ So this novel has a total of $44$or $43$pages.