1. Masha has an integer number of times more toys than Lena, and Lena has the same number of times more toys than Katya. Masha gave Lena 3 toys, and Katya gave Lena 2 toys. After that, the number of toys the girls had formed an arithmetic progression. How many toys did each girl originally have? Indicate the total number of toys the girls had.
1. Let initially Katya has $a$ toys, then Lena has $k a$, and Masha has $\kappa^{2} a$, where $\kappa-$ is a natural number $\geq 2$. After the changes: Katya has $a-2$, Lena has $a \kappa+5$, and Masha has $a \kappa^{2}-3$. Then these numbers form an arithmetic progression in some order. Let's consider the possible cases. If $a \kappa^{2}-3$ is the middle number in the arithmetic progression, then $2 a \kappa^{2}-6=a-2+a \kappa+5 \Rightarrow a\left(2 \kappa^{2}-\kappa-1\right)=9$. But $\kappa \geq 2$, so $2 \kappa^{2}-\kappa-1 \geq 5$ and therefore the possible equality: $2 \kappa^{2}-\kappa-1=9$. This equation does not have natural positive integer solutions. If $a \kappa+5$ is the middle number ($a-2$ is obviously not the middle number), then $2 a \kappa+10=a-2+a \kappa^{2}-3$. From this, $a(k-1)^{2}=15$. Therefore, $(\kappa-1)^{2} \in\{1,3,5,15\}$. If $(\kappa-1)^{2}=1$, then $\kappa=2, a=15 \Rightarrow 15,30,60$ are the required numbers. If $(\kappa-1)^{2}=3$, then $\kappa-$ is not an integer. The same statement is true in the other cases. Thus, Katya has 15 toys, Lena has 30, and Masha has 60. Therefore, the answer is: $\{105\}$