Eight children divided 32 peaches among themselves as follows. Anya received 1 peach, Katya - 2, Liza - 3, and Dasha - 4. Kolya Ivanov took as many peaches as his sister, Petya Grishin got twice as many peaches as his sister, Tolya Andreev - three times as many as his sister, and finally, Vasya Sergeev received four times as many peaches as his sister. Name the surnames of the four girls.
Try to represent the condition of the problem as a system of equations, and use the fact that the number of peaches received by each child is an integer.

## Solution

Let the number of peaches received by Kolya Ivanov's sister be denoted by $a$, by Petya Grishin's sister by $b$, by Tolya Andreev's sister by $c$, and by Vasya Sergeev's sister by $d$. Then the girls received $a, b, c$, and $d$ peaches, and the boys received $a, 2b, 3c$, and $4d$ peaches respectively. The numbers $a, b, c, d$ take the values $1, 2, 3$, and $4$, and $a + b + c + d = 10$. Additionally, $(a + b + c + d) + (a + 2b + 3c + 4d) = 32$. Therefore, $b + 2c + 3d = 12$.

Thus, $b$ and $d$ must have the same parity. If $b$ and $d$ are 1 and 3, then either $1 + 2c + 9 = 12$ or $3 + 2c + 3 = 12$. Both of these cases are impossible.

If $b$ and $d$ are 2 and 4, then either $2 + 2c + 12 = 12$ or $4 + 2c + 6 = 12$. The first of these cases is impossible, while in the second case we get $c = 1$. We then determine the values of the other unknowns: $b = 4, d = 2, a = 3$. Therefore, Kolya Ivanov's sister received 3 peaches, Petya Grishin's sister received 4, Tolya Andreev's sister received 1 peach, and Vasya Sergeev's sister received 2.

## Problem