Warm up problem 3: integer solutions

$(a + c)(b + d) = 18$ (*)

$18$ can be written as $1 \times 18, 2 \times 9$ or $3 \times 6$

$1 \times 18$ won't help, because $[(a + c), (b + d)] \geq 2$ , since $a, b, c$ and $d$ are positive integers.

Case $1 \rightarrow 2 \times 9$

There is only $1$ way for one of the factors of (*) to be $2$ and $8$ ways for the other one to be $9$ . Also, we can swap their values, doubling the number of solutions of this case. So this case gives us $1 \times 8 \times 2 = \boxed{16}$ solutions.

Case $2 \rightarrow 3 \times 6$

There are $2$ ways for one of the factors of (*) to be $3$ and $5$ ways for the other one to be $6$ . As with the first case, we can swap their values, so this case gives us $2 \times 5 \times 2 = \boxed{20}$ solutions.

Number of solutions: $16 + 20 = \boxed{36}$