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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?

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    For sure you can start saying that $40 \leq a + b + c + d \leq 80$. – 2017-01-02
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    @N.S.JOHN If $a$, $b$, $c$ and $d$ are all in the range $[10, 20]$... – 2017-01-02
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    You went wrong. Your solutions only satisfy the equation $(a-c)(b-d)= 168$ . They do not individually satisfy the given conditions – 2017-01-02
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    @N.S.JOHN Yes, I wasn't trying to do if and only if. – 2017-01-02

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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: $$b+d = 28\quad\implies\quad a+b+c+d = 26 + 28 = 54$$