$11 \cdot 11$ Let $n$ be a composite number, and $p$ be a proper divisor of $n$. Try to find the smallest positive integer $N$ (expressed in binary), such that $\left(1+2^{p}+2^{n-p}\right) N-1$ is divisible by $2^{n}$.
(31st International Mathematical Olympiad Preliminary Question, 1990)
[Solution] Clearly, the smallest positive integer $N$ that satisfies the problem's requirements should satisfy
$$
\left(1+2^{p}+2^{n-p}\right) N \equiv 1 \quad\left(\bmod 2^{n}\right) .
$$

Since $p$ is a proper divisor of $n$, then $p>1$, and thus
$$
\left(1+2^{p}+2^{n-p}, \quad 2^{n}\right)=1 \text {. }
$$

Therefore, the natural number $N$ that satisfies (1) and is less than $2^{n}$ is unique, and this $N$ is the number we are looking for.
It is not difficult to see that (1) is equivalent to
$$
\left(1+2^{p}+2^{n-p}+2^{n}\right) N \equiv 1 \quad\left(\bmod 2^{n}\right),
$$

which is
$$
\left(1+2^{p}\right)\left(1+2^{n-p}\right) N \equiv 1 \quad\left(\bmod 2^{n}\right) \text {. }
$$

Let $n=m p, \quad m>1$, then
$$
\left(1+2^{p}\right) \sum_{k=0}^{m-1}(-1)^{k} 2^{k p} \equiv 1-(-1)^{m} 2^{n} \equiv 1 \quad\left(\bmod 2^{n}\right) .
$$

On the other hand, since $n-2 p \geqslant 0$, we have
$$
\begin{aligned}
& \left(1+2^{n-p}\right)\left(1-2^{n-p}\right) \\
\equiv & 1-2^{n} 2^{n-2 p} \\
\equiv & 1 \quad\left(\bmod 2^{n}\right) .
\end{aligned}
$$

Thus,
$$
\left(1+2^{p}\right)\left(1+2^{n-p}\right)\left(1-2^{n-p}\right) \sum_{k=0}^{m-1}(-1)^{k} 2^{k p} \equiv 1 \quad\left(\bmod 2^{n}\right) .
$$

Comparing with (2), we have
$$
\begin{aligned}
N & \equiv\left(1-2^{n-p}\right) \sum_{k=0}^{m-1}(-1)^{k} 2^{k p} \\
& \equiv \sum_{k=0}^{m-1}(-1)^{k} 2^{k p}-2^{n-p} \\
& \equiv 2^{n}-2^{n-p}+\sum_{k=0}^{m-1}(-1)^{k} 2^{k p} \quad\left(\bmod 2^{n}\right) .
\end{aligned}
$$

Thus, we can let
$$
N=2^{n}-2^{n-p}+\sum_{k=0}^{m-1}(-1)^{k} 2^{k p} .
$$

Then $\quad 1 \leqslant N \leqslant 2^{n}$.
Therefore, $N$ is the number we are looking for.
Now, we write the binary representation of $N$ in two cases.
When $2 \nmid m$, from (3) we get

When 2 । $m$, from (3) we get