Find those consecutive integers whose sum is 100.
Solution. Consecutive integers form an arithmetic sequence where the difference is 1. Let the first term of the sought sequence be denoted by $a$, and the number of terms by $n$. According to the known sum formula:

$$
100=\frac{n}{2}(a+a+(n-1) 1)
$$

Rearranging, we get $n^{2}+2 a n-n=200$, that is (factoring out $n$) 200 must be factored into two factors:

$$
n(n+2 a-1)=200
$$

If $n=1$, then the sequence has only one term, which is 100.

The parity of $n$ and $n+2 a-1$ is different, so we only need to examine the factorizations of 200 where one factor is even and the other is odd. These are: $8 \cdot 25, 40 \cdot 5$, and $200 \cdot 1$. The last case for $n=1$ has already been examined.

If $n=8$, then $n+2 a-1=25$ and $a=9$,

$n=40$, then $n+2 a-1=5$ and $a=-17$,

$n=200$, then $n+2 a-1=1$ and $a=-99$,

$n=25$, then $n+2 a-1=8$ and $a=-8$,

$n=5$, then $n+2 a-1=40$ and $a=18$.

The following sequences satisfy the condition:

$$
\begin{array}{ll}
100 & n=1 \\
18,19,20,21,22 & n=5 \\
9,10,11, \ldots, 16 & n=8 \\
-8,-7,-6, \ldots, 16 & n=25 \\
-17,-16,-15, \ldots, 22 & n=40 \\
-99,-98,-97, \ldots, 100 & n=200
\end{array}
$$