Let's determine all arithmetic sequences consisting of natural numbers, with at most 240 terms, whose first term is a single-digit number, and whose first common term is 100, and another common term is 3103.
Let the first term of the sought arithmetic sequence be $a_{1}$ and the difference be $d$. Since the difference between any two terms of an arithmetic sequence is always divisible by the difference, $d$ can only be a divisor of

$$
3103-100=3003=3 \cdot 7 \cdot 11 \cdot 13
$$

However, $d$ cannot be $1,3,7,11$, because if $a_{1}$ is a single-digit number, even in the case of $d=11$, the 240th term $a_{1}+239 \cdot 11$ is still less than 3103; but $d$ cannot be greater than 100 either, because then 100 could not be a term of the sequence. Thus, the possible values of $d$ are: 13, $21,33,39,77,91$. Dividing 100 by these possible differences, the remainders provide the first terms. These are: $9,16,1,22,23,9$.

Since the first term is a single-digit number, only the sequences with differences of 13, 33, and 91 and the corresponding first terms of $9,1,9$ can be considered.

According to the example text, the first common term of the sequences is 100 (and not 9!), so the first and second, as well as the second and third sequences, represent a pair of solutions each.

$9,22,35,48,61,74,87,100, \ldots 3103$

$1,34,67,100, \ldots 3103 \ldots$

$9,100, \ldots 3103, \ldots$

János Kákóczy (Sátoraljaújhely, Kossuth g. IV. o. t.)