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On the table, there are 365 cards, with different numbers written on their reverse sides. For one ruble, Vasya can choose three cards and ask Petya to place them from left to right in ascending order of the numbers on the cards. Can Vasya, by spending 2000 rubles, guarantee to lay out all 365 cards on the table from left to right in ascending order of the numbers on them?

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Lemma. Suppose Vasya managed to lay out some $N-1$ cards in the required order on the table for $x$ rubles, where $N \leq 3^{k}$. Then he can add one more card to the laid-out cards, spending no more than $k$ rubles. We will prove this by induction on $k$. The base case $(k=1)$ is obvious.

Inductive step. Suppose $N=3 m-r \leq 3^{k}(r=0,1,2)$. Vasya will spend the first ruble to correctly lay out the $N$-th card and the cards $A$ and $B$, which are already in positions $m$ and $2 m$. Then the $N$-th card will fall into one of the three parts into which the other two cards divide the already laid-out sequence. But the cards $A$ and $B$ divide the sequence of $N-1$ cards into pieces of no more than

$m-1 \leq 3^{k-1}-1$ cards, so Vasya has to determine the position of the card with number $N$ among no more than $3^{k-1}-1$ cards, spending no more than

$k-1$ rubles, which can be done by the inductive hypothesis.

Now, using the lemma, let's calculate Vasya's expenses for laying out all 365 cards. Vasya will spend 1 ruble on laying out the first three cards. To add cards numbered from 4 to 9 (a total of 6 cards), Vasya will spend no more than 2 rubles on each. For cards numbered from 10 to 27 - no more than 3 rubles on each, from 28 to 81 - no more than 4 rubles, from 82 to 243 - no more than 5 rubles, and finally, for cards numbered from 244 to 365 - no more than 6 rubles on each.

In total, Vasya will be able to lay out all the cards on the table in the required order, spending no more than $1+6 \cdot 2+18 \cdot 3+$ $54 \cdot 4+162 \cdot 5+122 \cdot 6=1845$ rubles.

## Answer

Yes.