A father gives 11 apples to one son, 18 to another, and 25 to the third. How is it possible that each of them brought home the same amount of money, even though they sold the apples at the same price?
First solution. First, 1 f is given for 4 drbot, so A sells 8 drbot, B sells 16 drbot, and C sells 24 drbot; then the rest are sold for 2 f each. In this case, A received $2+6=8 \mathrm{f}$, B received $4+4=8 \mathrm{f}$, and C received $6+2=8 \mathrm{f}$.

(Cukor Gábor, Budapest, ag.hitv.ev. fogymn.)

Second solution. First, 8 apples were sold for $x$ fillér; A sold 8 apples, B sold 16, and C sold 24. Then the rest were sold at $x$ fillér each, so

$$
A x+3 x=4 x \mathrm{f}-\mathrm{t}, \quad B 2 x+2 x=4 x \mathrm{f}-\mathrm{t}, \quad \text { and } \quad C 3 x+x=4 x \text { f-t received. }
$$

(Fried Ernö, Budapest, V. district.)

The problem was also solved by: Babocsai Gy., Bendl K., Dénes M., Kirchknopf E., Koffler B., Kovács Gy., Miklóssy K.