# Problem 1.

B-1

Find $x$ :

$$
\frac{1}{2+\frac{1}{3+\frac{1}{4+\frac{1}{5+\frac{1}{x}}}}}=\frac{16}{37}
$$
Answer: $-0.25$

Solution. $\frac{1}{2+\frac{1}{3+\frac{1}{4+\frac{1}{5+\frac{1}{x}}}}}=\frac{16}{37} \Leftrightarrow 2+\frac{1}{3+\frac{1}{4+\frac{1}{5+\frac{1}{x}}}}=\frac{37}{16} \Leftrightarrow \frac{1}{3+\frac{1}{4+\frac{1}{5+\frac{1}{x}}}}=\frac{5}{16} \Leftrightarrow 3+\frac{1}{4+\frac{1}{5+\frac{1}{x}}}=\frac{16}{5} \Leftrightarrow$ $\Leftrightarrow \frac{1}{4+\frac{1}{5+\frac{1}{x}}}=\frac{1}{5} \Leftrightarrow \frac{1}{5+\frac{1}{x}}=1 \Leftrightarrow \frac{1}{x}=-4 \Leftrightarrow x=-\frac{1}{4}$.

B-2

Find $x$ :

$$
\frac{1}{6+\frac{1}{7+\frac{1}{8+\frac{1}{9+\frac{1}{x}}}}}=\frac{64}{393}
$$

Answer: $-0.125$

B-3

Find $x$ :

$$
\frac{1}{5+\frac{1}{4+\frac{1}{3+\frac{1}{2+\frac{1}{1+\frac{1}{x}}}}}}=\frac{17}{89}
$$

Answer: $-0.5$

B-4

Find $x$ :

$$
\frac{1}{4+\frac{1}{5+\frac{1}{6+\frac{1}{7+\frac{1}{x}}}}}=\frac{67}{281}
$$

Answer: $-0.2$

## Lomonosov Olympiad for Schoolchildren in Mathematics

Preliminary Stage 2020/21 Academic Year for $7-8$ grades

## B-1

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