10-3-1. Non-negative integers $a, b, c, d$ are such that

$$
a b+b c+c d+d a=707
$$

What is the smallest value that the sum $a+b+c+d$ can take?
Answer: 108.

Solution variant 1. The given equality can be rewritten as

\[
(a+c)(b+d)=7 \cdot 101
\]

where the numbers 7 and 101 are prime. Therefore, either one of the expressions in parentheses is 1 and the other is 707, or one of the expressions in parentheses is 7 and the other is 101. In the first case, \(a+b+c+d\) equals 708, and in the second case, it equals 108, and 108 is the smaller of these two numbers.

Comment. It follows from the solution that the equality given in the problem is indeed possible. We can take any non-negative integers with the conditions \(a+c=7\), \(b+d=101\). For example, \(a=1\), \(c=6\), \(b=1\), \(d=100\).