Task 1. For the natural numbers $a, b, c, d$ it is known that

$$
a c+a d+b c+d b=68 \text { and } c+d=4
$$

Calculate the value of the expression $a+b+c+d$.
Solution. From $a c+a d+b c+d b=68$ we obtain

$$
a \cdot(c+d)+b \cdot(c+d)=68
$$

that is,

$$
(a+b) \cdot(c+d)=68
$$

From $c+d=4$, we get $4(a+b)=68$, hence, $a+b=17$. Therefore, $a+b+c+d=17+4=21$.