Let's find the four-digit number $\overline{x y z u}$ written in the decimal system, for which

$$
u+z-4 x=1 \ldots
$$

$$
\text{and } u+10 z-2 y=14 \ldots
$$
Given $u \leq 9$ and $z \leq 9$, from 1)

$$
u+z=4 x+1 \leq 18 \ldots
$$

The numbers of the form $4 x+1$ that are less than 18 are: $1, 5, 9, 13, 17$, i.e., $x=0,1,2,3,4$. From 2)

$$
u+10 z=14+2 y \leq 32
$$

since $y \leq 9$. However, if

$$
u+10 z \leq 32 \ldots
$$

then only $z=1,2,3$ are possible. The case $z=0$ can be excluded because then from 2)

$$
u=14+2 y>9
$$

which is impossible. Based on this:

If $z=1$, then from 1) $u=4 x$, so $x=1, 2$ and $u=4, 8$, and $y=0, 2^{2}$

If $z=2$, then from 1) $u=4 x-1$, so $x=1, 2$ and $u=3, 7$; with these values, we do not get an integer for $y$. ${ }^{2}$

If $z=3$, then from 1) $u=4 x-2$, so $x=1, 2$ and $u=2, 6$. With these, from 2) $y=9, 11$. The latter is not possible. According to the requirements of the problem, 3 numbers meet the criteria:

$1014, \quad 2218, \quad 1932$.

János Füsz (Cistercian Saint Stephen High School, Székesfehérvár).[^0]


[^0]:    ${ }^{1} x=0$ cannot be considered, because then we would not get a four-digit number.

    $\left.{ }^{2} \right)$ from 2) $2 y=u+10 z-14$.