The sum of three numbers is 35. If we increase the third number by 1, we get an arithmetic sequence, and if we increase the first number by 3, we get a geometric sequence. Which are these three numbers?
Let the three numbers be $x, y, z$. According to the problem, $x, y$, and $z+1$ form an arithmetic sequence, thus

$$
2 y=x+z+1
$$

Furthermore, $x+3, y, z$ form a geometric sequence, which means

$$
y^{2}=z(x+3)
$$

The third equation of our system is

$$
x+y+z=35
$$

From (1), we have $x+z=2 y-1$, substituting this into (3) gives

$$
3 y-1=35, \quad \text { hence } \quad y=12
$$

Substituting this value of $y$ into (3) and (2) respectively, we get

$$
\begin{gathered}
x+z=23 \\
z(x+3)=144
\end{gathered}
$$

Substituting the value of $z$ from (4) into (5) gives

$$
(23-x)(x+3)=144
$$

which simplifies to

$$
x^{2}-20 x+75=0
$$

from which

$$
x_{1}=15, \quad x_{2}=5
$$

and thus

$$
z_{1}=8, \quad z_{2}=18
$$

Therefore, two triplets of numbers satisfy the conditions:

$$
15,12,8 \quad \text { and } \quad 5,12,18 \text {. }
$$

Gábor Szatmári (Bp. VIII., Piarist g. II. o. t.)