Example 4. Find the length of the common chord of the circles $x^{2}+y^{2}-10 x-10 y=0$ and $x^{2}+y^{2}+6 x+2 y-40=0$.
Subtracting the two equations yields $4 x+3 y-10=0$. (Equation of the common chord)

Completing the square for $x^{2}+y^{2}-10 x-10 y=0$, we get $(x-5)^{2}+(y-5)^{2}=50$.
Thus, the center of the circle is $\mathbf{C}(5,5)$.
The distance from the center of the circle to the common chord is
$$
\mathrm{d}=\frac{|20+15-10|}{5}=5 \text {. }
$$

By the Pythagorean theorem, the length of the chord is
$$
\mathrm{l}=2 \sqrt{50-25}=10 \text {. }
$$

$2^{n}$ cannot be divisible by 32. However, by assumption, $2^{n}(n \geqslant 6)$ can be divided by $2^{6}=64$, leading to a contradiction.

Assume $2^{n}(n \geqslant 6)$ meets the problem's requirements and ends with the digit 2. In this case, the last few digits of $2^{\mathbf{n}}$ could be 02, 22, 012, or it could be of the form
$$
2^{\mathbf{n}}=10 \underbrace{0 \cdots 0112}_{\mathrm{m} \text { digits }} .(\mathrm{m} \geqslant 0)
$$

Since 02 and 22 are not divisible by 4, and 012 is not divisible by 8, in these cases, $2^{n}$ cannot be divisible by 64. Therefore, the first two scenarios are impossible. For the last scenario, it is also impossible. Because $\left.1112=2^{3} \cdot 1306 \mathrm{~m}=0\right)$, $10112=2^{7} \cdot 79(\mathrm{~m}=1)$, it is clear that they are not powers of 2. When $\mathrm{m} \geqslant 2$, $10^{\mathrm{m}+3}$ is divisible by 32, but 112 is not divisible by 32. This also leads to a contradiction. In summary, only $n=5$ meets the requirements.