Let's determine the multiples of 5 that, when divided by 11, leave a remainder of 4.
I. Solution. If two numbers, when divided by 11, give the same remainder, then their difference is a multiple of 11. However, these two numbers are also multiples of 5, so their difference is as well. Therefore, the difference between the numbers in question is a multiple of 55, since 11 and 5 are prime numbers. This implies that if $N$ is a number that meets the conditions, then

$$
N+55m
$$

also meets the conditions. Since 15 is a suitable number, the numbers we are looking for are generally of the form $15+55m$, that is:

$15,70,125,180, \ldots$

Imre Böröcz (Cistercian St. Imre g. VI. o. Bp. XI.).

II. Solution. The numbers $N$ that, when divided by 11, give a remainder of 4, can be written in the form $11x+4$. Furthermore, it is required that

$$
\frac{11x+4}{5}=2x+\frac{x+4}{5}
$$

be an integer. This occurs if $\frac{x+4}{5}$ is an integer, i.e.,

$$
x+4=5t, \quad x=5t-4
$$

and thus

$$
N=11(5t-4)+4=55t-40
$$

where $t=1,2,3,4,5 \ldots$

László Huhn (Kegyesrendi g. V. o. Szeged.).