A mother distributes walnuts from a basket among her 4 children. Tamás receives one walnut and a quarter of the remaining walnuts; then Erzsi receives one walnut and a quarter of the remaining walnuts; Béla also receives one walnut and a quarter of the remaining walnuts; finally, Juliska receives one walnut and a quarter of the remaining walnuts. It turned out that the boys received 100 more walnuts than the girls. How many walnuts were there originally, and how many were left in the basket at the end?
I. solution: Let's assume that there were $x$ walnuts in the basket before the distribution began.

According to the problem:

Tamás gets $1+\frac{x-1}{4}=\frac{x+3}{4}$ walnuts, and $\frac{3 x-3}{4}$ remain.

Erzsi gets $1+\frac{1}{4}\left(\frac{3 x-3}{4}-1\right)=\frac{16}{16}+\frac{3 x-7}{16}=\frac{3 x+9}{16}=\frac{3}{4} \cdot \frac{x+3}{4}$ walnuts,

$$
\text { and } x-\frac{7}{4} \cdot \frac{x+3}{4}=\frac{9 x-21}{16} \text { remain. }
$$

Béla gets $1+\frac{1}{4}\left(\frac{9 x-21}{16}-1\right)=\frac{9 x+27}{64}=\left(\frac{3}{4}\right)^{2} \frac{x+3}{4}$,

$$
\text { and } x-\frac{37}{6} \cdot \frac{x+3}{4}=\frac{27 x-111}{64} \text { remain. }
$$

Juliska gets $1+\frac{1}{4}\left(\frac{27 x-111}{64}-1\right)=\frac{27 x+81}{256}=\left(\frac{3}{4}\right)^{3} \frac{x+3}{4}$.

Let $\frac{x+3}{4}=z$, then according to the problem

$$
z+\frac{9 z}{16}=\frac{3 z}{4}+\frac{27 z}{64}+100
$$

from which

$$
z=\frac{x+3}{4}=256
$$

and thus

$$
x=1021
$$

The distributed walnuts form a geometric sequence of four terms, whose first term is $z=256$, and the common ratio is $\frac{3}{4}$, so their sum is

$$
s_{4}=256 \frac{1-\frac{81}{256}}{1-\frac{3}{4}}=256 \cdot 4 \cdot \frac{175}{256}=700
$$

and thus in the basket there remain

$$
1021-700=321 \text { walnuts. }
$$

Ferenc Kolonits (Bp., VIII., Piarist g. I. o. t.)

II. solution: If Tamás got $(1+a)$ walnuts, then $3 a$ walnuts remained. Erzsi got from this $1+\frac{3 a-1}{4}=\frac{3}{4}(a+1)$. With the same reasoning, we find that Béla got three-quarters of Erzsi's walnuts, and Juliska got three-quarters of Béla's walnuts. Thus, Erzsi got three-quarters of Tamás's walnuts, Béla got $\frac{9}{16}$ of Tamás's walnuts, and Juliska got $\frac{27}{64}$ of Tamás's walnuts.

Since

$$
\left(1+\frac{9}{16}\right)-\left(\frac{3}{4}+\frac{27}{64}\right)=\frac{25}{64}
$$

the boys got $\frac{25}{64}$ of Tamás's walnuts more than the girls, which according to the problem is 100 walnuts. Therefore, Tamás got $100: \frac{25}{64}=256$ walnuts. But Tamás got $1+a$ walnuts, and thus originally there were $4 a+1=4 \cdot 255+1=1021$ walnuts.

Juliska got $\left(\frac{3}{4}\right)^{3} 256=\frac{27}{81} \cdot 256=108$ walnuts, and in the basket there remained $3(108-1)=321$ walnuts.

Kálmán Győry (Ózd, József A. g. II. o. t.)