A four-digit (integer) number, when divided by 7 and 29, gives a remainder of 0, and if its 19 times multiple is divided by 37, the remainder is 3. What is this number?
The sought number $x$ is a four-digit number, hence

$$
1000 \leq x \leq 9999
$$

Furthermore, it is divisible by the relatively prime numbers 7 and 29, so it is a multiple of $7 \cdot 29=203$: $x=203 y$, where $y$ is a natural number, thus

$$
1000 \leq 203 y \leq 9999, \text { from which } 5 \leq y \leq 49
$$

Finally, $19 x-3$ is divisible by 37, so

$$
19 x-3=19 \cdot 203 y-3=37 z
$$

where $z$ is a natural number. Let's determine $z$ first:

$$
z=\frac{3857 y-3}{37}=104 y+\frac{9 y-3}{37}=104 y+u
$$

where $u=(9 y-3) / 37$ is a natural number. From this:

$$
y=\frac{37 u+3}{9}=4 u+\frac{u+3}{9}=4 u+v
$$

where $v=(u+3) / 9$ is a natural number, and finally from this $u=9 v-3$.

Thus $y=4(9 v-3)+v=37 v-12, \quad x=203 y=7511 v-2436$. Substituting this into (1) and rearranging,

$$
1000 \leq 7511 v-2436 \leq 9999, \quad 3436 \leq 7511 v=12435
$$

This is only satisfied by $v=1$, and with that $x=7511-2436=5075$. Indeed, $5075=7 \cdot 725=29$ and $19 \cdot 5075=96425=$ $37 \cdot 2606+3$.

Iren Faludi (Bp. VIII., Ságvári E. tutor. Ig. I1. o. t.)

Note: We can reach the goal a bit faster this way: $u=(9 y-3) / 37=3(3 y-1) / 37$ is only an integer if $3 y-1=37 w$, so with the smallest and largest values of $y$ according to (2) $3 \cdot 5-1=14 \leq 37 w \leq 146=3 \cdot 49-1$, i.e., $1 \leq w \leq 3$. On the other hand, $y=12 w+(w+1) / 3$ is only an integer with these values of $w$ for $w=2$: $y=25, x=203 \cdot 25=5075$.

Klara Bornes (Bp. XIV., Teleki Blanka lg. II. o. t.)