The 24th All-Union Mathematical Olympiad has a problem:

There are 1990 piles of stones, with the number of stones being $1, 2, \cdots$, 1990. The operation is as follows: each time, you can choose any number of piles and take the same number of stones from each of them. How many operations are needed at least to take all the stones away?
Solve the confusion of
$$
\begin{array}{l}
1990=: 2^{10}+2^{9}+2^{8}+2^{7}+2^{6}+0 \cdot 2^{5} \\
+0 \cdot 2^{4}+0 \cdot 2^{3}+2^{2}+2^{1}+0 \cdot 2^{0},
\end{array}
$$

and write $1,2, \cdots, 1989$ in binary form. The operation is as follows:

First, take away $2^{10}=1024$ stones from each pile that has enough; second, take away $2^{9}=$ 512 stones from each pile that has enough; …… finally, take away the stones from the piles that have only $2^{0}=1$ stone left. This way, a total of 11 operations are used.
Since there are piles with only one stone, there must be an operation to take one stone; if the remaining operations all take more than 2 stones, then the piles with exactly two stones cannot be taken away. Therefore, 2 operations can take at most $1+2=3$ stones, ..., 10 operations can take at most $1+2+$ $2^{2}+\cdots+2^{9}=1023$ stones. Thus, 1950 piles of stones require at least 11 operations.