3. proposed by A. Shen

There is an infinite one-way strip of cells, numbered with natural numbers, and a bag with ten stones. Initially, there are no stones in the cells of the strip. The following actions are allowed:

- moving a stone from the bag to the first cell of the strip or back;
- if there is a stone in the cell with number $i$, then a stone can be moved from the bag to the cell with number $i+1$ or back.

Is it possible, acting according to these rules, to place a stone in the cell with number $1000 ?$
Answer: $\partial a$.

Solution. Note that for each action there is an inverse to it. Therefore, if we get from situation $A$ to situation $B$ by following the rules, we can also get from situation $B$ to situation $A$ by following the rules.

We will show by induction that if there is a reserve of $n$ stones, then, acting according to the above rules, we can place a stone in any cell from 1 to $2^{n}-1$.

Base of induction: $n=1$. This is obvious.

Inductive step. Suppose we have proven that with a reserve of $n$ stones, we can place a stone in all cells up to the $\left(2^{n}-1\right)$-th. Now let's assume we have $n$ black stones and one red stone. We will act as follows:

(1) Without removing the red stone from the bag, place a black stone in the $\left(2^{n}-1\right)$-th cell. This can be done by the inductive hypothesis.

(2) Place the red stone in the $2^{n}$-th cell.

(3) Perform the operations as in step (1), but in reverse and in the opposite order. It is clear that the red stone will not hinder this. In the end, all black stones will be back in the bag, and on the strip, there will be exactly one stone - the red stone in cell $2^{n}$. Let's agree that we will not remove the red stone from now on.

(4) The cells with numbers from $2^{n}+1$ to $2^{n+1}-1$ form a strip of length $2^{n}-1$. The inductive hypothesis for $n$ stones applies to this strip, as the red stone allows operations with the leftmost cell of this strip. Therefore, we can place a stone in the last cell.

Thus, with a reserve of 10 stones, we can place a stone in all cells numbered from 1 to $1023=2^{10}-1$.

Comment: In fact, a strip of length 1000 is sufficient. Indeed, note that among the cells numbered from 1000 to 1023, the first stone will be placed in cell 1000. That is, to place a stone in the 1000th cell, cells numbered from 1001 to 1023 are not necessary, which means that with a reserve of 10 stones, we can place a stone in the last cell of a strip of length 1000.