Shen A.H.

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 can be performed:

- move a stone from the bag to the first cell of the strip or vice versa;
- if there is a stone in the cell with number \$\$, then a stone can be moved from the bag to the cell with number $\$ \mathrm{i}+1 \$$ or vice versa.

Is it possible, acting according to these rules, to place a stone in the cell with number 1000?
Note that for every action, there is an inverse action. Therefore, if we move from situation \$A\$, acting according to the rules, to situation \$ \mathrm{~B} \$$, then from situation \$ \mathrm{~B} \$$ we can return to situation \$ \mathrm{~A} \$$ by acting according to the rules.

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

Base case: \$ 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 \$ (2 \wedge n-1) \$$-th cell. Now suppose we have \$ n \$$ black stones and one red stone. We will act as follows:

1) Without removing the red stone from the bag, we will ensure that the \$ (2 \wedge n-1) \$$-th cell contains a black stone. This can be done by the inductive hypothesis.
2) Place the red stone in the \$ 2 \wedge n \$$-th cell.
3) Perform the operations as in step (1), but in the opposite and reverse order. It is clear that the red stone will not interfere with 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 the \$ 2 \wedge n \$$-th cell. We agree that we will not remove the red stone from now on.
4) The cells with numbers from \$ 2 \wedge n+1 \$$ to \$ 2 \wedge\{n+1\}-1 \$$ form a strip of length \$ 2 \wedge\{n\}-1 \$$. The inductive hypothesis applies to \$ n \$$ stones for 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=$ $2 \wedge\{10\}-1 \$$.

## Answer

Yes.