Ostrovsky $M$.

A number from 1 to 144 is guessed. You are allowed to select one subset of the set of numbers from 1 to 144 and ask whether the guessed number belongs to it. You have to pay 2 rubles for a "yes" answer and 1 ruble for a "no" answer. What is the minimum amount of money needed to surely guess the number?
Let $a_{1}=2, a_{2}=3, a_{i}=a_{i-} 1+a_{i-} 2$ for $i \geq 3$. Then $a 10=144$. We will prove by induction that among not less than $a_{i}$ numbers, the guessed number cannot be found by paying less than $i+1$ rubles.

For $i=1$ and $i=2$, this is true.

Suppose there are not less than $a_{i}$ numbers. Then either the set $M$ of numbers highlighted in the first question contains not less than $a_{i-} 2$ numbers (first case), or the set of numbers not in $M$ contains not less than $a_{i-1} 1$ numbers (second case). In the first case, if the guessed number is in $M$, then 2 rubles need to be paid for the answer, and, by the induction hypothesis, at least ( $i-2$ )+1 rubles are needed to guess the number, i.e., a total of at least $i+1$ rubles.

In the second case, if the guessed number is not in $M$, then 1 ruble needs to be paid for the answer and at least ( $i$ 1)+1 ruble for guessing the number, i.e., again a total of at least $i+1$ rubles.

The algorithm for guessing the number is clear from the previous reasoning: at each step, the set $M$ of $a_{i}$ numbers containing the guessed number should be divided into sets $M_{1}$ of $a_{i-} 2$ numbers and $M_{2}$ of $a_{i-} 1$ numbers, and ask whether the number belongs to the set $M_{1}$.

## Answer

11 rubles.