Number Theory Problem No.2

Since brute force is to be used, we must well have computer doing the work. Using the Code environment of Brilliant.org:

By using logic and brute force, we have 9 trios: $(45,4,0), (42,14,9), (41,18,6), (40,21,0), (39,22,6), (39,18,14), (36,27,4), (36,24,13), (32,24,21).$

Now it's time for combinatorics.

Since the problem asked for the integer solutions, we have to include both positive- and negative-containing solutions. Of the given result, we see that there are 7 all-nonzero trios, which can be shuffled (For example: $(45,4,0)$ and $(4,45,0)$ or $(0,4,45)$ are different) in $3!$ ways and have $2^{3}$ combinations of positive or negative value for each trios.

For the remaining two, we also have $3!$ ways to shuffle but only $2^{2}$ positive-or-negative-value combinations since $0$ is a neutral number (it's neither positive nor negative).

The answer is $7 \times 3! \times 2^{3} + 2 \times 3! \times 2^{2} = \boxed{384}$ different solutions.