4. In the multi-story fraction

$$
D+\frac{1}{P+\frac{1}{0+\frac{1}{b+\frac{1}{b}}}}
$$

the letters were replaced with non-zero digits. What is the smallest result that can be obtained? (As usual, different letters correspond to different digits).
Answer. $1+\frac{1}{9+\frac{1}{2+\frac{1}{8+\frac{1}{3}}}}=1 \frac{53}{502}$.

Solution. Let the entire fraction be $T$ and write

$$
\mathrm{T}=\text { D }+\frac{1}{X}, \quad X=\mathrm{P}+\frac{1}{Y}, \quad Y=\mathrm{O}+\frac{1}{\mathrm{Z}}, \quad Z=\mathrm{B}+\frac{1}{\mathrm{~b}}
$$

Notice that all numbers $\mathrm{X}, \mathrm{Y}, \mathrm{Z}$ are greater than 1, so the second terms in the first three expressions are less than 1. If we change the first term in them, it will change by at least 1, meaning the sum will change in the same direction (the second term can change by less than 1 and will not compensate for the change in the first term). Therefore, $T$ is smaller the smaller D is, so we should take $D=1$. Also, $T$ is smaller the larger $X$ is, so $P$ should be as large as possible, i.e., $P=9$. The number $X$ is larger the smaller $Y$ is, which means the smaller O is. Therefore, $\mathrm{O}=2$. Meanwhile, $Y$ is smaller the larger $Z$ is. The maximum $Z$ is achieved with the maximum $B=8$ and the minimum $b=3$, from which we get the answer.

Criteria. Full solution - 7 points. Correct answer without justification - 2 points.