Between 100 and 1000, which numbers are divisible by 7 and, when divided by either 4 or 9, leave a remainder of 3?
The least common multiple of 4 and 9 is 36, so the sought number can be written in the form $36 y + 3$, or equivalently $7 z$; thus

$$
x = 36 y + 3 = 7 z,
$$

from which

$$
z = \frac{36 y + 3}{7} = 5 y + \frac{y + 3}{7} = 5 y + u
$$

so

$$
y = 7 u - 3
$$

and

$$
z = 36 u - 15
$$

and thus

$$
x = 7 z = 252 u - 105
$$

if we substitute 1, 2, 3, 4 for $u$, then the values of $x$ are

$147, 399, 651, 903$.

(Imre Kürti, Eger.)

The problem was also solved by: Baranyó A., Bartók I., Demjén E., Deutsch I., Enyedi B., Füstös P., Glück J., Haar A., Kertész G., Kiss J., Liebner A., Ligeti P., Pám M., Pichler S., Pivnyik I., Riesz K., Riesz M., Schlesinger O., Schuster Gy., Schwemmer I., Szántó H., Szücs A., Tóth B., Weisz P.