Problem 10.3. On the board, natural numbers $a, b, c, d$ are written. It is known that among the six sums

$$
a+b, \quad b+c, c+d, d+a, a+c, b+d
$$

three are equal to 23, and the other three are equal to 34.

(a) (1 point) What is the value of $a+b+c+d$?

(b) (3 points) What is the smallest of the numbers $a, b, c, d$?
Answer: (a) 57. (b) 6.

Solution. (a) Let's add all 6 sums $a+b, b+c, c+d, d+a, a+c, b+d$. Since three of them are equal to 23 and the other three are equal to 34, we get $23 \cdot 3 + 34 \cdot 3$. On the other hand, we get $3(a+b+c+d)$. Therefore,

$$
a+b+c+d=\frac{23 \cdot 3 + 34 \cdot 3}{3}=57
$$

(b) Suppose that among the numbers $a, b, c, d$ there are at least three different ones. Without loss of generality, let's assume these are $a, b$, and $c$, with $a < b < c$. Then $a+b < a+c < b+c$, which contradicts the problem's condition.

Suppose all numbers $a, b, c, d$ are the same. Then all their pairwise sums are also the same, which again contradicts the problem's condition.

Therefore, among the numbers $a, b, c, d$ there are exactly two different ones. Without loss of generality, we only need to consider three cases:

- $a=b=c<d$
- $a=b<c=d$
- $a<b=c=d$.

Case 1. Let $a=b=c<d$. Then

$$
2a = a+b = a+c = b+c < a+d = b+d = c+d
$$

We get the system

$$
\left\{\begin{array}{l}
2a = 23 \\
a+d = 34
\end{array}\right.
$$

but it is not solvable in natural numbers.

Case 2. Let $a=b<c=d$. Then $a+b < b+c < c+d$, which contradicts the problem's condition.

Case 3. Let $a<b=c=d$. Then

$$
a+b = a+c = a+d < b+d = b+c = c+d = 2d
$$

We get the system

$$
\left\{\begin{array}{l}
a+d = 23 \\
2d = 34
\end{array}\right.
$$

the solution of which is $a=6, b=c=d=17$. Therefore, the smallest of the numbers is 6.