The following puzzle is one of three deductive-reasoning problems (‘brain-teasers’) that were choosable to attempt for an assignment. Although it posits that "only simple arithmetic and logic" are required to solve it, I suspect that some particular math skills might help greatly.
What is an optimal method to approach it and satisfactorily find all the solution[s] (in a manner showing that the answer is correct with no missing solutions)?
An initial deduction shows that 9 can belong to only one string-combination (9,3,1), which means that it cannot be a string connector and that (9,3,1) exists which leaves only two other odd letters (5, 7) from the set of 9, and writing out the isolated possibilities for all others 8 thru 1 (although doing so is more exhaustive than an efficient method) yields the number of possible respective combinations being 8=two, 7=two, 6=two, 5=three, 4=four, 3=three, 2=three, 1=three (all more than one). Also observable is that 13 being odd must consist of an odd number of odds (#= 3 or 1) and an even number of evens (#= 0 or 2), which restricts the possibilities of potential configurations of the entire sequence.
Add $(A+B+C)+(C+D+E)+(E+F+G)+(G+H+I)=52$.
$A+B+C+D+E+F+G+H+I=45$ so we have $C+E+G=7$
$C,E,G$ must be $1,2,4$ in some order.
The $1$ must be part of two of $1+4+8$, $1+3+9$, and $1+5+7$. We will rule out $1+5+7$ in a bit. As you said, $9$ must only be used once. If it is part of one of the middle runs, $3$ would be used twice, but that is not allowed, so $9$ is part of an end run. The other end of the other run from $1$ has to be a turn, so it has to be $2$ or $4$. It cannot be $1+5+7$. We can start (or end) with either $3,9$ or $9,3$. Say we start with $9,3$. Then the next run is $1,8,4$ so we have a proper turn at the end. The other turn is $2$. The number between them must be $7$.
One solution is $9,3,1,8,4,7,2,5,6$. Each end pair may be interchanged, giving four solutions. We can reverse any of them, giving a total of eight solutions.