(solved by Alice Héliou). The sum of twenty consecutive integers is 1030. What is the smallest of these integers?
. If $n$ denotes the smallest of these integers, the sum is:

$$
n+(n+1)+(n+2)+\cdots+(n+19)=20 n+190
$$

We are thus reduced to solving the equation $20 n+190=1030$, which gives $n=42$.