6. There are four children in the family. They are 5, 8, 13, and 15 years old. The children are named Anya, Borya, Vера, and Galya. How old is each child, if one girl attends a kindergarten, Anya is older than Borya, and the sum of Anya's and Vера's ages is divisible by 3?
6. First solution method. The problem is about four children (Anya, Borya, Vera, and Galya) and four ages of children (5 years, 8 years, 13 years, and 15 years).

Let's represent the problem's condition in a table.

| Anya | Borya | Vera | Galya |
| :---: | :---: | :---: | :---: |
| $5,8,13,15$ | $5,8,13,15$ | $5,8,13,15$ | $5,8,13,15$ |

Since one girl attends a kindergarten and Anya is older than Borya, it follows that Anya and Borya cannot be 5 years old.

| Anya | Borya | Vera | Galya |
| :---: | :---: | :---: | :---: |
| $8,13,15$ | $8,13,15$ | $5,8,13,15$ | $5,8,13,15$ |

According to the condition, the sum of Anya's and Vera's ages is divisible by 3. From this, the girls can be 13 and 5 years old or 8 and 13 years old. But neither of them can be 15 years old.

| Anya | Borya | Vera | Galya |
| :---: | :---: | :---: | :---: |
| 8,13 | $8,13,15$ | $5,8,13$ | $5,8,13,15$ |

Since Anya is older than Borya, she can only be 13 years old, Borya 8 years old, and Vera 5 years old. In this case, Galya will be 15 years old.

Second solution method. We can reason differently. Since one of the girls attends a kindergarten, Borya cannot be 5 years old. According to the condition, Anya is older than Borya. Therefore, she can be 13 or 15 years old. According to the third condition, the sum of Anya's and Vera's ages is divisible by 3. But only the sum of 13 and 5 is divisible by 3. Thus, Anya is 13 years old, Vera is 5 years old, Galya is 15 years old. Therefore, Borya is 8 years old.

Answer: Vera is 5 years old, Borya is 8 years old, Anya is 13 years old, Galya is 15 years old.