2. Petya wrote down five consecutive terms of an arithmetic progression and encrypted them according to the principle: each digit was replaced with a letter, different digits with different letters, and vice versa. Here's what he got: A, BV, BS, VD, EE. What numbers did Petya write?
Answer: $5,12,19,26,33$.

Solution. From the condition, it follows that the terms of the arithmetic sequence are natural numbers and the difference of the progression, denoted as $d$, is positive. We consider the letters A, B, V, C, D, E as digits or numbers, and the corresponding two-digit numbers will be denoted as $\overline{\mathrm{BB}}$ and so on.

Since the second and third terms of the progression start with the same digit, the difference of the progression is less than 10. Therefore, $\mathrm{E}=1, \mathrm{~B}=2, \mathrm{E}=3$. Taking into account the already found digits, the terms of the sequence can be written as: $\mathrm{A}, 12,1 \overline{\mathrm{C}}, 2 \bar{Д}, 33$. From this, $33-12=21=3 d$ and, therefore, the terms of the sequence are $5,12,19,26,33$. These numbers satisfy the condition of the problem.