In the Library of Count Dracula Am thinking about this question from the book Analysis $1$ by Herbert Amann and Joachim Escher:
In the library of Count Dracula no two books contain exactly the same number of
words. The number of books is greater than the total number of words in all the books.
These statements suffice to determine the content of at least one book in Count Dracula’s
library. What is in this book?
My reasoning: If there is a total of $n$ words in all books, then there can be at most $n$ different books (since each book must have a different number of words). So how can there be more books than the total number of words in all books?
Am assuming there are no books with $0$ words.
 A: I was reading this in Amann's Analysis 1 last night (I know, strange bedtime reading). I thought I would flesh out my take on it since the question is still unanswered (even if the comments state the solution).
You are forced to assume that one book has only blank pages:
Let $n$ be the number of books and $x_i$ the number of words in the $i$-th book. I interpret "The number of books is greater than the total number of words in all the books" as
$$
x_1 + \cdots + x_n < n
$$
If $x_i > 0$ for all $i$ you get
$$
  x_1 + \cdots + x_n \geq 1 + \cdots + 1 = n
$$
which contradicts the previous inequality. So at least one of the $x_i$ has to be $0$.
Now, if you bring in the additional assumption $x_i\neq x_j$ for $i\neq j$, you can prove that the only options are $n=1, x_1=0$, or $n=2, x_1 = 0, x_2=1$:
You have $n$ non-negative numbers $x_1,\dots, x_n$ which are all different. Also $x_i < n$ for all $i$ otherwise the sum would be $\geq n$. Since there are only $n$ integers between $0$ and $n-1$, you have, up to a relabeling of the books, $x_i = i - 1$. But then
$$
  x_1 + \cdots + x_n = 0 + 1 + \cdots + (n-1) = \frac{(n-1)n}{2}< n
$$
which reduces to
$$
  n(n - 3) < 0
$$
and the only solutions for $n$ an integer are $n = 1$ and $n = 2$.
