Problem 1.1. Recent years can be expressed as sums, subtractions, and multiplications of numbers with the same and unique digit; for example:

$$
2009=7 \times 7 \times 7 \times 7-7 \times 7 \times 7-7 \times 7, \quad 2010=66 \times 6 \times 6-66 \times 6+6 \times 6-6
$$

Can the same be done with 2011, without ever repeating the same addend? For example, $2011=1+1+1+\ldots$ is not allowed.
Solution Problem 1.1 If 2011 could be expressed as sums, subtractions, and multiplications of numbers with the same digit $a$, since each of these numbers is divisible by $a$, it follows that $a$ is a divisor of 2011. Now, 2011 is a prime number, so $a=1$.

It is easy to observe that

$$
\begin{aligned}
1000 & =1111-111 \\
2 & =111-11 \times 11+11+1
\end{aligned}
$$

Multiplying these two equalities, we get:

$$
\begin{aligned}
& 2000= \\
& 1111 \times 111-1111 \times 11 \times 11+1111 \times 11+1111-111 \times 111+ \\
& +111 \times 11 \times 11-111 \times 11-111
\end{aligned}
$$

Verify that all the addends are distinct from each other and different from 11. Therefore, by adding 11 to the number above, we get a solution.

There are infinitely many different ways:

$$
2011=1111 \times 1111-111 \times 11111-111 \times 111+111 \times 11 \times 11-111+11+1
$$

or

$$
2011=1111 \times 1111-111 \times 11111+1111-111+11
$$