We can change a natural number $n$ in three ways:

a) If the number $n$ has at least two digits, we erase the last digit and we subtract that digit from the remaining number (for example, from 123 we get $12-3=9$ );

b) If the last digit is different from 0, we can change the order of the digits in the opposite one (for example, from 123 we get 321 );

c) We can multiply the number $n$ by a number from the set $\{1,2,3, \ldots, 2010\}$.

Can we get the number 21062011 from the number 1012011?
The answer is NO. We will prove that if the first number is divisible by 11, then all the numbers which we can get from $n$, are divisible by 11. When we use $a$), from the number $10 a+b$, we will get the number $m=a-b=11 a-n$, so $11 \mid m$ since $11 \mid n$. It's well-known that a number is divisible by 11 if and only if the difference between sum of digits on even places and sum of digits on odd places is divisible by 11. Hence, when we use $b$), from a number which is divisible by 11, we will get a number which is also divisible by 11. When we use $c$), the obtained number remains divisible by 11. So, the answer is NO since 1012011 is divisible by 11 and 21062011 is not.