In what base number systems can a three-digit number be equal to twice the number formed by reversing the order of its digits?
Let the digits of a number in base $g$ be denoted by $a, b$, and $c$. According to the given condition,

$$
a g^{2}+b g+c=2 c g^{2}+2 b g+2 a
$$

and it is evident that the inequality $a>c>0$ must hold. Rearrange the equation to the following form:

$$
(a-2 c) g^{2}=b g+(2 a-c)
$$

From the inequality $a>c$, it follows that $2 a-c>0$. Since $b$ is a digit, we have $0 \leq b \leq g-1$. Therefore, the right-hand side of the equation is a positive number. The maximum value of the right-hand side is less than $g^{2}+g$ due to $b \leq g-1$ and $2 a-c<2 g$. Since the equality can only hold if the left-hand side is also positive, and its smallest value is $g^{2}$, we have

$$
(a-2 c) g^{2}=b g+(2 a-c)=g^{2}
$$

From this, it follows that $a-2 c=1, b=g-1$ and $2 a-c=g$. Solving the system of equations for $a$ and $c$, we get

$$
a=\frac{2 g-1}{3} \quad \text { and } \quad c=\frac{g-2}{3} .
$$

$c$ can only be a positive integer if $g=3 k+2$, where $k$ is a positive integer. In this case, $a$ is also an integer; $a=2 k+1$ and $b=3 k+1, c=k$. These numbers satisfy the requirements of the problem.

In summary, a three-digit number can equal twice the number formed by reversing its digits in those number systems where the base of the number system gives a remainder of 2 when divided by 3. In the binary number system, a solution is only possible if we allow the first digit of the reversed number to be 0: $110=2 \cdot 011$.