2. Gena encrypted a sequence of five integers: B, AN, AX, NO, FF. Here, different digits are denoted by different letters, and the same digits by the same letters; a comma separates adjacent numbers (and in “NO” the second character is the letter “O”). Gena forgot the encrypted numbers but remembers that the difference between any two adjacent numbers is the same (the right number minus the left number). Can the numbers Gena encrypted be uniquely determined?
Answer: $5,12,19,26,33$.

Solution. All these numbers can be determined if we know the first number and the difference $d$ between two adjacent numbers. The first digits of the second and third numbers coincide, meaning they are in the same decade, and their difference, equal to $d$, does not exceed 9. Therefore, by adding $d$ to the first (single-digit) number, we can only get a two-digit number starting with 1, that is, $\mathrm{A}=$ 1. Similarly, $\mathrm{H}=2, \Phi=3$, because in these cases, a number not exceeding 9 is added to a two-digit number, and the first digit increases by 1. We get the sequence: Б, 12, 1Х, $2 \mathrm{O}, 33$.

Notice that $1 \mathrm{X}-12=2 \mathrm{O}-1 \mathrm{X}=33-2 \mathrm{O}=d$. Adding these equations, we get: $33-12=3 d$, from which $d=7$. We have restored the last number and the difference. It is now easy to restore the sequence: $5,12,19,26,33$.

Criteria. 1 point for finding each of the digits A, H, Ф. 3 points for finding the difference $d$, and 1 point for the final answer, points are summed. If the correct answer is found by trial and the uniqueness is not proven - 5 points. If the sequence: Б, $12,1 \mathrm{X}, 2 \mathrm{O}, 33$ is justified, but trial is used further - 6 points.