Example 8 Find a three-digit number in decimal, which has the following property: when its digits are written in the original order in some number system other than the decimal system, the number is twice the original number.

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Example 8 Find a three-digit number in decimal, which has the following property: when its digits are written in the original order in some number system other than the decimal system, the number is twice the original number.
Solution: Let the hundreds digit, tens digit, and units digit of the required number be $x, y, z$, and let $c$ be the base of the new number system. According to the problem, we have
$$
2(100 x+10 y+z)=c^{2} x+c y+z
$$

which simplifies to $\left(200-c^{2}\right) x+(20-c) y+z=0$.
Thus, the original problem is transformed into finding integer solutions $x, y, z, c$ for equation (3) that satisfy the following conditions:
$$
1 \leqslant x \leqslant 9,0 \leqslant y \leqslant 9,0 \leqslant z \leqslant 9, c>x, c>y, c>z
$$

It is easy to find that $c=15$.
This is because when $00$, and it is not equal to 0;
When $c \geqslant 16$, the left side of (3) has
$\left(200-c^{2}\right) x+(20-c) y+z \leqslant-56 x+4 y+z \leqslant-56+36+9=$ $-11<0$, which is also not equal to 0, hence $c=15$.
Thus, equation (3) becomes $5 y+z=25 x$.
Clearly, $5 \mid z$, so $z=0,5$.
When $z=0$, $y=5 x$, combining with (4) we get $x=1, y=5$, so the required three-digit number is 150, and we have $(150)_{15}=1 \times 15^{2}+5 \times 15=2 \times 150$.

When $z=5$, $y+1=5 x$, combining with (4) we get $x=1, y=4$, or $x=2, y=9$, thus we obtain the three-digit numbers 145, 295, and we have
$$
\begin{array}{l}
(145)_{15}=15^{2}+4 \times 15+5=2 \times 145, \\
(295)_{15}=2 \times 15^{2}+9 \times 15+5=2 \times 295 .
\end{array}
$$

Therefore, the three-digit numbers that satisfy the problem are: 150, 145, and 295.