The pages of a notebook are numbered consecutively such that the first sheet has the numbers 1 and 2, the second sheet has the numbers 3 and 4, and so on. A sheet is torn out of this notebook. The page numbers of the remaining sheets are added up, resulting in a sum of 2021.

(a) How many pages could the notebook originally have had?

(b) Which page numbers could be on the torn-out sheet?

(Walther Janous)

Answer. There is exactly one solution. The notebook had 64 pages and the sheet with the page numbers 29 and 30 was torn out.
We denote the number of leaves by $b>0$. Then the number of pages is $2b$ (and in particular even). We are looking for a number $2b$ such that

$$
1+2+\ldots+(2b-1)+2b=\frac{(2b) \cdot (2b+1)}{2}>2021
$$

holds. Since $\frac{60^2}{2}=1800$ is roughly the appropriate order of magnitude, we obtain through systematic trial:

$$
\frac{62 \cdot 63}{2}=1953<2021
$$

and

$$
\frac{64 \cdot 65}{2}=2080>2021
$$

This means that the remaining sum is certainly greater than or equal to 2080 with at least 33 leaves. Therefore, there can be no solution with more than 32 leaves.

(Lukas Donner)