10. (20 points) Given various natural numbers $a, b, c, d$, for which the following conditions are satisfied: $a>d, a b=c d$ and $a+b+c+d=a c$. Find the sum of all four numbers.
Answer: 12

Solution. From the relations $a>d$ and $ab=cd$, we have the inequality $ba+d$ and $\frac{ac}{2} \geqslant 2c > b+c$. Adding them, we get a contradiction with the condition.

Therefore, without loss of generality, assume that $a \geqslant 3$. Since $a$ is a natural number, then $a \in\{1,2,3\}$. Moreover, $d$ is a natural number less than $a$, so $a \neq 1$. Let's consider two cases.

Case 1. $a=3$. The relations can be rewritten as

$$
3b=cd, 3+b+c+d=3c, 3>d
$$

From the last inequality, in particular, it follows that $(3, d)=1$. Therefore, $3 \mid c$. Since all numbers are distinct, $c \geqslant 6$. Then we have a chain of inequalities

$$
3+b+c+d < 3c \Rightarrow b+c+d < 2c \Rightarrow b+d < c
$$

Since $3>d$ implies $d=1$. The relations can be rewritten as $2b=c, 3+b=c$. The only solution to this system of equations is the pair $(b, c)=(3,6)$. We find the solution $(2,3,6,1)$, the sum of which is 12.