Problem 2. Misha has a set of nine cards with the letters of the word "LOMOMONOSOV". On the reverse side of each card, Misha wrote a digit such that the digits on cards with the same letters are the same, and the digits on cards with different letters are different. It turned out that the equation

$$
L+\frac{O}{M}+O+H+\frac{O}{C}=OB
$$

is true, where both fractions in the equation are proper. What digits could Misha have written on the cards? It is sufficient to provide one example.
Answer: $8+\frac{2}{3}+2+9+\frac{2}{6}=20$ (the 8 and 9, as well as 3 and 6, can be swapped).

Solution. The left side of the resulting correct equality is less than the value of $10+9+10=29$, which means the digit corresponding to the letter "O" can be either 1 or 2. In this case, the equality $\frac{\mathrm{O}}{\mathrm{M}}+\frac{\mathrm{O}}{\mathrm{C}}=1$ must hold. From this, it follows that the digit on the card with the letter "O" can only be 2, which means the digits on the cards with the letters "M" and "C" are (in any order) 3 and 6. In this case, the inequality $L +\mathrm{H}+1+2 \geqslant 20$ holds, from which the values of the digits on the remaining cards are restored.