2. In the country of Limpopo, 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 points in the country) are given below:

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The number on the arrow indicates how many units, specified at the end of the arrow, can be obtained for one unit, specified 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 $1 / 15$ dollars. (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 $\$ \leftrightarrows$ B are prohibited in Limpopo.

Money can be transported across the border of Limpopo only in dollars. Uncle Vasya arrived in Limpopo with 100 dollars. He can perform the above exchange operations any number of times, but has no other sources of income. Can he get rich and take 200 dollars out of Limpopo? If yes, explain how. If not, prove it.
# Answer. It can.

Solution. Note that the problem statement does not explicitly state whether it is required to obtain no less than $200 \$$, or exactly $200 \$$$. The answer is positive for both interpretations. Although it is clear that a solution for the case "exactly $200 \$$" also solves the second case, we will provide different solutions for both cases, as both interpretations are often found in participants' work.

a) Need to obtain no less than $200 \$$.

Consider the cycle $\mathrm{K} \rightarrow \mathcal{E} \rightarrow$ B $\rightarrow \mathrm{K}$. If money is exchanged according to this cycle, the amount of money increases. Indeed, if there were initially $x$ coconuts, they are exchanged for $\frac{x}{11}$ raccoons, which in turn are exchanged for $\frac{6 x}{11}$ bananas, which are then exchanged for $\frac{12 x}{11}$ coconuts. Thus, from $x$ coconuts, $\frac{12 x}{11}$ coconuts are obtained.

Uncle Vasya's strategy is as follows: initially exchange 100 dollars for 1000 coconuts. Then exchange them according to the cycle mentioned above until more than 3000 coconuts are obtained. For this, the cycle will need to be repeated $n$ times, where $n$ is such that $\left(\frac{12}{11}\right)^{n}>3$. Since $\frac{12}{11}>1$, this inequality is equivalent to the inequality $n>\log _{\frac{12}{11}} 3$. Clearly, a natural number $n$ satisfying this condition exists.

The resulting coconuts should be exchanged for dollars, of which there will be no less than $\frac{3000}{15}=200$.

b) Need to obtain exactly $200 \$$$. Consider the same cycle as in part a), but each time run the exchange cycle with exactly 11 coconuts, i.e., each time set aside the profitable 12th coconut. Then, after $n$ cycles, we can earn exactly $n$ coconuts. Now the strategy, for example, is as follows: exchange $100 \$$ for 1000 coconuts, take 11 out of these 1000 and perform the cycle $n$ times, resulting in $1000+n$ coconuts. If $n=2000$, then in the end we will get 3000 coconuts, which we will exchange back for $200 \$$.

Criteria.

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

(+/-) An algorithm that provides exponential capital growth has been provided, but there is no proof that this algorithm will ever reach the required amount, or an incorrect proof is provided, for example, the unboundedness of capital is "derived" from its growth.

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