4. In the country of Lumumba, there are four national currencies: bananas (B), coconuts (K), raccoons $(\mathcal{E})$, and dollars (\$). The exchange rates for these currencies (the same in all exchange offices in the country) are given below:

$$
\mathrm{B} \underset{\frac{1}{2}}{\stackrel{2}{\rightleftarrows}} \mathrm{K} \quad \mathcal{E} \underset{\frac{1}{6}}{\stackrel{6}{\longrightarrow}} \mathrm{K} \quad \mathcal{E} \underset{\frac{1}{11}}{\stackrel{11}{\rightleftarrows}} \mathrm{K} \quad \$ \underset{\frac{1}{15}}{\stackrel{10}{\rightleftarrows}} \mathrm{K}
$$

The number on the arrow indicates how many units of the currency at the end of the arrow can be obtained for one unit of the currency at the beginning of the arrow. For example, one raccoon can be exchanged for 6 bananas or 11 coconuts, one dollar for 10 coconuts, and one coconut for $1 / 15$ dollar. (When solving the problem, any currency can be divided into arbitrarily small parts: for example, $101 / 43$ raccoons can be exchanged for $606 / 43$ bananas). Exchanges $\$ \leftrightarrows \mathcal{E}$ and $\$ \leftrightarrows$ in Lumumba are prohibited.

Money can be transported across the border of Lumumba only in dollars. Uncle Vasya arrived in Lumumba with 100 dollars. He can exchange any currency for any other currency an unlimited number of times. Can he get rich and take 200 dollars out of Lumumba using only currency exchange operations? If yes, explain how. If no, prove it.
Solution. Consider the cycle $\mathrm{K} \rightarrow \mathcal{E} \rightarrow$ B $\rightarrow$ K. This cycle allows increasing the amount of currency. Indeed, if initially there were 11 coconuts, they can be exchanged for 1 raccoon, which in turn can be exchanged for 6 bananas, which can be exchanged for 12 coconuts. Thus, from 11 coconuts, we get 12 coconuts.

We will run this cycle exactly 11 coconuts each time, i.e., we will set aside the profitable 12th coconut each time. Then, after $n$ cycles, we can earn exactly $n$ coconuts. Now, for example, the strategy is: exchange $100 \$$ for 1000 coconuts, take 11 out of these 1000 and perform the cycle $n$ times, resulting in $1000+n$ coconuts. If $n \geq 2000$, then in the end we will get $\geq 3000$ coconuts, which we can exchange back for $\geq 200 \$$.

Criteria.

(-/+) A cycle that increases capital has been found, and no other significant progress has been made.

(+/-) An algorithm using multiplication by $\frac{12}{11}$ at each cycle is provided, but it is not proven that any amount can be achieved this way. That is, it is not proven that $\left(\frac{12}{11}\right)^{n}$ can become greater than 3, or an incorrect proof is provided (for example, the inequality $\left(\frac{12}{11}\right)^{n}>3$ is derived from the increase of the fraction $\left(\frac{12}{11}\right)^{n}$ as $n$ increases).

(+) The problem is fully solved. In the case of providing an algorithm with linear capital growth (similar to the one provided above), justification that the required amount will be reached is not required.