## Solution.

Notice that with the move

$$
n=(n-1)+1 \quad \mapsto \quad(n-1) \cdot 1=n-1
$$

the spoken number decreases exactly by 1,

so by a sequence of such moves, the spoken number can be decreased at will.

With the move

$$
n=(n-2)+2 \quad \mapsto \quad(n-2) \cdot 2=2 n-4
$$

the spoken number $n$ increases if $2 n-4>n$, i.e., if $n>4$.

Starting from any number $n>4$, by repeating that move, we can reach some number greater than 2011,

and then by decreasing as described above, we can obtain the number 2011.

Notice that the numbers $1,2,3,4$ cannot be increased with allowed moves:

$$
\begin{array}{llll}
2=1+1 & 3=2+1 & 4=3+1 & 4=2+2 \\
1 \cdot 1=1 & 2 \cdot 1=2 & 3 \cdot 1=3 & 2 \cdot 2=2
\end{array}
$$

so starting from them, we cannot reach the number 2011.

Therefore, the game could have started with any number $n$ that satisfies $n \geqslant 5$.
## Task A-1.5.

Friends Anica and Neda are playing a game where, in each move, after one of them says a number $n$, the other must say a number of the form $a \cdot b$ where $a$ and $b$ are natural numbers such that $a + b = n$. The game then continues in the same way, from the number just spoken. With which numbers could the game have started if, after some time, one of them said the number 2011?