6. How many solutions does the rebus $\frac{B+O+C+b+M+O+\breve{U}}{K+J+A+C+C}=\frac{22}{29}$ have, where different letters represent different digits, and the same letters represent the same digits? It is known that the digit 0 is not used.
Solution. In the rebus, 9 different letters are used, i.e., all non-zero digits are used. $1+2+3+0+0 \leq K+Л+A+C+C \leq 6+7+8+9+9$, i.e.

$$
6 \leq K+Л+A+C+C \leq 39 . \quad \text { Since the fraction } \frac{22}{29} \text { is irreducible, the number }
$$

$K+Л+A+C+C$ must be divisible by 29, the only case that satisfies all conditions is $K+Л+A+C+C=29$. Therefore, $B+O+C+b+M+O+\breve{И}=22$, in the word, 6 different letters are used (O is repeated twice). Note that $1+2+3+4+5+6=21$, doubling 1 gives 22, so $\mathrm{O}=1$. The unused digits form the word КЛА, so $7+8+9+\mathrm{C}+\mathrm{C}=29,2 \mathrm{C}=5$, but C cannot be non-integer. There are no solutions.

## Answer 0

## Grading Criteria.

Correct answer obtained with justification - 7 points.

Correctly shown that the numerator is 22 and the denominator is 29 without further progress - $\underline{2}$ points.

Shown that O must be equal to 1 - $\underline{\mathbf{1} \text { point. }}$

Underlined points are summed.

Incorrect solution or only the answer - 0 points.