4. 220 There is a group of children, the sum of their ages is 50 years, the oldest is 13 years old, and one is 10 years old; except for the 10-year-old child, the ages of the remaining children form an arithmetic sequence. How many children are there? How old is each child?
［Solution］ Let the age of the oldest child be $a$, i.e., $a=13$.
Suppose, apart from the 10-year-old child, there are $b+1$ children, and let the common difference of their ages be $d$. Then the ages of these $b+1$ children are
$$
a, a-d, a-2 d, \cdots, a-b d
$$

Thus, $a+(a-d)+(a-2 d)+\cdots+(a-b d)$
$$
=(b+1) a-\frac{b(b+1)}{2} d=50-10=40,
$$

which means $(b+1)(2 a-b d)=80$.
It is clear that $(b+1)$ divides 80, but $2 a-b d=a+(a-b d)>13$, so $b+1<\frac{80}{26}$. Considering both aspects, we know that $b+1$ can only be 4 or 5.

When $b+1=4$, $4(26-3 d)=80$, hence $d=2$. Substituting the values of $b$ and $d$ into (1) gives: $13,11,10,9,7$. Therefore, there are $4+1=5$ children in total, and their ages are $13,11,10,9,7$.
When $b+1=5$, $5(26-4 d)=80$, in this case $d$ is not an integer, so this situation is impossible. Thus, the solution is unique.