Let's find all three-digit numbers in the decimal system that, when considered as a number in a larger base, give us twice the original number.
Let the sought digits be $X, Y, Z$, and the base of the number system that meets the requirement be $B$. Then, by rearranging,

$$
\begin{gathered}
2(100 X+10 Y+Z)=X \cdot B^{2}+Y B+Z \\
\left(200-B^{2}\right) \cdot X+(20-B) \cdot Y+Z=0
\end{gathered}
$$

This is not the case for $B=15$, because then on the one hand,

$$
\left(200-B^{2}\right) X \leq(200-256) X \leq-56
$$

since $X \geq 1$, and on the other hand,

$$
(20-B) Y+Z<5 \cdot 9+9=54
$$

thus the left side of (1) would be negative.

If $B=15$, then (1) becomes:

$$
-25 X+5 Y+Z=0, \quad Z=5(5 X-Y)
$$

so the value of $Z$ is 0 or 5. In the first case, $5 X-Y=0,5 X=Y<10$, so $X=1, Y=5$. In the second case, $5 X-Y=1$, which yields two solutions: $X=1, Y=4$ and $X=2, Y=9$.

Thus, the three-digit number in question can only be 150, 145, and 290. When considered as numbers in a base-15 number system, they indeed represent a number that is twice as large as if the base were 10.

Nagy Zsigmond (Budapest, Kaffka M. g. III. o. t.)