A poor woman brought a basket of eggs to the market. Someone bumped into her, causing her to drop the basket and all the eggs broke; however, the person responsible, wishing to compensate for the damage, inquired about the number of eggs; the woman replied that she did not know for sure, but she remembered well that when she arranged them in pairs, one remained; the same happened when she arranged them in threes, fours, fives, or sixes, but when she counted them in sevens, none remained. The question is, could you figure out the number of eggs from this information?

[^0]:    ${ }^{1 *}$ Submitted by: György Lukácsi, high school teacher, N.-Bánya; Antal Novák, high school teacher, Sz.-Ujvár; Aladár Visnya, Budapest.
2, 3, 4, 5 and 6 have the least common multiple of 60; therefore, the number we are looking for, $x$, must be expressible in the form $60 y+1$ or $7 z$; that is,

$$
x=60 y+1=7 z
$$

from which

$$
z=8 y+\frac{4 y+1}{7}=8 y+u
$$

where

$$
7 u=4 y+1
$$

and thus

$$
y=u+\frac{3 u-1}{4}=u+t
$$

where

$$
4 t=3 u-1
$$

from which

$$
u=t+\frac{t+1}{3}=t+s
$$

where

$$
3 s=t+1
$$

and thus

$$
t=3 s-1
$$

Substituting the value of $t$ into $u$, $u$ and $t$ into $y$, and finally $y$ into $x$:

$$
x=420 s-119
$$

where $s$ is any positive integer; some values of $x$ are: 301, 721, 1141, etc.

(Frigyes Riesz.)

Number of solutions: 36.