11.2. On the board, the expression $\frac{a}{b} \cdot \frac{c}{d} \cdot \frac{e}{f}$ is written, where $a, b, c, d, e, f$ are natural numbers. If the number $a$ is increased by 1, then the value of this expression increases by 3. If in the original expression the number $c$ is increased by 1, then its value increases by 4; if in the original expression the number $e$ is increased by 1, then its value increases by 5. What is the smallest value that the product $b d f$ can have?

(N. Agakhanov)
# Answer: 60.

First solution. Let the value of the original expression be $A$. Then, as a result of the first operation, the product will take the value $\frac{a+1}{a} \cdot A=A+3$, from which $A=3a$. This means that $A$ is a natural number. Moreover, from this equality, it follows that it is divisible by 3. Similarly, it is proven that the number $A$ is divisible by 4 and by 5, and that $A=4c=5e$. From the pairwise coprimality of the numbers 3, 4, and 5, it follows that $A$ is divisible by $3 \cdot 4 \cdot 5=60$. Therefore, $A \geqslant 60$.

Rewriting the equality $\frac{a}{b} \cdot \frac{c}{d} \cdot \frac{e}{f}=A$ as $\frac{A}{3b} \cdot \frac{A}{4d} \cdot \frac{A}{5f}=A$, we get $A^{2}=60bdf$, from which $bdf=\frac{A^{2}}{60} \geqslant 60$. It remains to provide an example showing that the product of the denominators can be equal to 60. One possible example is: $\frac{20}{3} \cdot \frac{15}{4} \cdot \frac{12}{5}$.

Second solution. As in the first solution, we get $A=3a=4c=5m$, from which

$$
\frac{1}{b} \cdot \frac{c}{d} \cdot \frac{e}{f}=3, \quad \frac{a}{b} \cdot \frac{1}{d} \cdot \frac{e}{f}=4, \quad \frac{a}{b} \cdot \frac{c}{d} \cdot \frac{1}{f}=5
$$

Multiplying the first equation by the second and dividing by the third, we get that $\frac{e^{2}}{bdf}=\frac{12}{5}$; since the fraction on the right is irreducible, the denominator $bdf$ is divisible by 5. Similarly, it is proven that it is divisible by 3 and by 4, from which it follows that it is divisible by 60, that is, it is not less than 60.

Comment. Only the answer - 0 points.

An example is provided where $bdf=60-1$ point.

It is only proven that $bdf \geqslant 60-4$ points.