Three numbers form a geometric progression; if we subtract 16 from the third number, we get an arithmetic progression. If we then subtract 2 from the second term of this arithmetic progression, we get a geometric progression again. What are these numbers?
Let the sought numbers be: $x, y, z$. According to the problem:

$$
\begin{gathered}
y^{2}=x z \\
2 y=x+z-16 \\
(y-2)^{2}=x(z-16)
\end{gathered}
$$

From (1), substituting the value of $y^{2}$ into (3):

$$
y=4 x+1
$$

Substituting this value of $y$ into (1) and (2), we get:

$$
16 x^{2}+8 x+1=x z
$$

$$
7 x+18=z
$$

Multiplying both sides of the equation (6) by $x$ and subtracting the resulting equation from (5), we get:

$$
9 x^{2}-10 x+1=0
$$

from which

$$
x_{1}=1 \text { and } x_{2}=\frac{1}{9} \text {. }
$$

From (4) and (2), we further obtain:

$$
y_{1}=5, y_{2}=\frac{13}{9} \text { and } z_{1}=25, z_{2}=\frac{169}{9} \text {. }
$$

(Goldstein Zsigmond, high school VIII. grade, Nyíregyháza.)

The problem was also solved by: Bálint Béla, Freund Antal, Friedmann Bernát, Geist Emil, Grünhut Béla, Hofbauer Ervin, Kántor Nándor, Klein Mór, Kornis Ödön, Porde Gyula, Riesz Frigyes, Schneider Béla, Szabó István, Weisz Ármin.