What is the value of $a+b+c+d$? 
Four integers $a,b,c,d$ make all the statements below true. What is the value of $a+b+c+d$?
(i) $10 \leq a,b,c,d \leq 20$
(ii) $ab-cd = 58$
(iii) $ad-bc = 110$

Adding the two equations together gives $ab+ad-cd-bc = 168$ and so $$a(b+d)-c(b+d) = (a-c)(b+d) = 168 = 2^3 \cdot 3 \cdot 7.$$ Then since $20 \leq b+d \leq 40$, we see that $b+d = 21,24,28$. In the first case, $b+d = 21$ and $a-c = 8$. Thus, $(a,c) = (18,10),(19,11)$ and $(b,d) = (10,11),(11,10)$. How do we deal with the other cases?

Consider $(iii)-(ii)$, we have
$$(a+c)(d-b) = (ad-bc) - (ab-cd) = 110 - 58 = 52$$
Since $10 \le a, c \le 20 \implies 20 \le a+c \le 40$ and the only divisor
of $52$ between $20$ and $40$ is $26$, we find
$$a+c = 26, d - b = 2$$
Consider $(ii)+(iii)$, we have
$$(a-c)(b+d) = (ab-cd) + (ad - bc) = 58+160 = 168$$
Since $10 \le b, d \le 20 \implies 20 \le b+d \le 40$ and the divisors of
$168$ between $20$ and $40$ are $21, 24, 28$, we have 3 possibilities.
$$
\begin{cases}
b+d = 21,a-c = 8\\
b+d = 24,a-c = 7\\
b+d = 28,a-c = 6
\end{cases}$$
Notice 
$$\begin{cases}
a+c = 26 &\implies a-c = 26 - 2c \equiv 0\pmod 2\\
d-b = 2  &\implies b+d = 2 + 2b \equiv 0 \pmod 2
\end{cases}
$$ This rules out the first and second case. This leaves us with only
one and only one possibility:
$$b+d = 28\quad\implies\quad a+b+c+d = 26 + 28 = 54$$
