11*(!). Given the sum: \(2^{8}+2^{7}+2^{6}+2^{5}+2^{4}+2^{3}+2^{2}+2+1\). Is it possible to obtain any natural number from 1 to 511 inclusive by crossing out some of the addends? Find a method (algorithm) to do this, if it exists, of course.
11. Possible. Solution. In this case, we will limit ourselves to a few examples to show how to do it.

\[
\begin{aligned}
& 405=2^{8}+149 \\
& 149=2^{7}+21 \\
& 21=2^{4}+5 \\
& 5=2^{2}+1 \\
& 405=2^{8}+2^{7}+2^{4}+2^{2}+1
\end{aligned}
\]

Similarly: \(57=2^{5}+2^{4}+2^{3}+1\). All calculations are simplified if you use a table of powers of 2.