
6. In a book with page numbers from 1 to 100 , some pages are torn off. The sum of the numbers on the remaining pages is 4949. How many pages are torn off?


Solution: Suppose $r$ pages of the book are torn off. Note that the page numbers on both the sides of a page are of the form $2 k-1$ and $2 k$, and their sum is $4 k-1$. The sum of the numbers on the torn pages must be of the form

$$
4 k_{1}-1+4 k_{2}-1+\cdots+4 k_{r}-1=4\left(k_{1}+k_{2}+\cdots+k_{r}\right)-r
$$

The sum of the numbers of all the pages in the untorn book is

$$
1+2+3+\cdots+100=5050
$$

Hence the sum of the numbers on the torn pages is

$$
5050-4949=101
$$

We therefore have

$$
4\left(k_{1}+k_{2}+\cdots+k_{r}\right)-r=101
$$

This shows that $r \equiv 3(\bmod 4)$. Thus $r=4 l+3$ for some $l \geq 0$.

Suppose $r \geq 7$, and suppose $k_{1}101
\end{aligned}
$$

Hence $r=3$. This leads to $k_{1}+k_{2}+k_{3}=26$ and one can choose distinct positive integers $k_{1}, k_{2}, k_{3}$ in several ways.

![](https://cdn.mathpix.com/cropped/2024_06_05_6f4d28fa51cab68f2750g-4.jpg?height=48&width=326&top_left_y=2072&top_left_x=868)

