11.36 Find a three-digit number that has the following property: when its digits are written in the same order in a non-decimal base, the number is twice the original number:
(Polish Mathematical Olympiad, 1961)
［Solution］Let the non-decimal numeral system be base $c$, and let the hundreds, tens, and units digits of this three-digit number in decimal be $x, y, z$ respectively.
From the problem, we have $2(100 x+10 y+z)=x \cdot c^{2}+y \cdot c+z$.
This can be rewritten as $\left(200-c^{2}\right) x+(20-c) y+z=0$.
Thus, the problem reduces to finding integer solutions $x, y, z, c$ for equation (1) that satisfy the following conditions:
$$
\begin{array}{l}
1 \leqslant x \leqslant 9, \quad 0 \leqslant y \leqslant 9, \quad 0 \leqslant z \leqslant 9, \\
c>x, \quad c>y, \quad c>z .
\end{array}
$$

We first prove that the base $c$ of the new numeral system can only be 15.
Indeed, if $0<c \leqslant 14$, then
$$
\left(200-c^{2}\right) x+(20-c) y+z \geqslant 4 x+6 y+z \geqslant 4,
$$
and equation (1) cannot hold.
If $c \geqslant 16$, then
$$
\begin{array}{c}
\left(200-c^{2}\right) x+(20-c) y+z \\
\leqslant-56 x+4 y+z \\
\leqslant-56+36+9 \\
=-11 .
\end{array}
$$

Thus,
$$
c=15 \text {. }
$$

Substituting $c=15$ into equation (1), we get
$$
-25 x+5 y+z=0 .
$$

If integers $x, y$, and $z$ satisfy equation (2), since $-25 x$ and $5 y$ are both multiples of 5, then $z$ must also be a multiple of 5. Therefore, $z=0$ or $z=5$.
If $z=0$, then equation (2) becomes
$$
\begin{array}{l}
-5 x+y=0, \\
x=1, y=5 .
\end{array}
$$

Thus, the only solution is
At this point, the three-digit number in decimal is 150, and in base 15, $(150)_{15}=15^{2}+5 \cdot 15=$ $300=2 \cdot 150$. This meets the problem's requirements.
If $z=5$, then equation (2) becomes
$$
-5 x+y+1=0 .
$$

Thus, when $x \geqslant 3$,
$$
-5 x+y+1 \leqslant-15+9+1=-5<0 .
$$

So, it must be that $x<3$, i.e., $x=1$ or $x=2$.
When $x=1$, $y=4$.
When $x=2$, $y=9$.
At this point, the three-digit numbers in decimal are 145 and 295.
In base 15, $(145)_{15}=15^{2}+4 \cdot 15+5=290=2 \cdot 145$.
$$
(295)_{15}=2 \cdot 15^{2}+9 \cdot 15+5=590=2 \cdot 295 .
$$

Thus, there are three solutions to this problem, which in decimal are 145, 150, and 295.