12.61 There is a group of children, the sum of whose ages is 50 years. The oldest is 13 years old, and one of them is 10 years old. Excluding the 10-year-old child, the ages of the remaining children form an arithmetic sequence. How many children are there, and how old is each child?
(China Beijing High School Mathematics Competition, 1956)
［Solution］ Let the age of the oldest child be $a$, i.e., $a=13$.
Suppose there are $b+1$ children excluding the 10-year-old child, and let the common difference of their ages be $d$. Then the ages of these $b+1$ children are
$$
a, a-d, a-2 d, \cdots, a-b d,
$$

and the sum of their ages is $50-10=40$. Therefore,
$$
\begin{array}{c}
a+(a-d)+\cdots+(a-b d)=40, \\
(b+1) a-\frac{b(b+1)}{2} d=40, \\
(b+1)(2 a-b d)=80 .
\end{array}
$$

Thus, $b+1$ must divide 80.
Also,
$$
2 a-b d=a+(a-b d)>13,
$$

so
$$
\begin{array}{l}
b+1>\frac{80}{26}
\end{array}
$$

which means
$$
\frac{80}{26}<b+1<\frac{80}{13} \text {. }
$$

Therefore, $b+1$ can only be 4, 5, or 6. Since 6 is not a divisor of 80, $b+1$ can only be 4 or 5.
When $b+1=4$, $\quad b=3$.
$$
(b+1)(2 a-b d)=4(26-3 d)=80 \text {. }
$$

So
$$
d=2 \text {. }
$$

Thus, $a-d=11, a-2 d=9, \quad a-3 d=7$.
Therefore, there are $4+1=5$ children in total, and their ages are $13,11,10,9,7$.
When $b+1=5$, $b=4$.
$$
(b+1)(2 a-b d)=5(26-4 d)=80,
$$

which gives $d=2.5$, which is not an integer.
So $b+1=5$ is not possible.
Thus, there is only one unique solution.