A company ordered a larger gift item. During delivery, two of them were away, so to balance the bill, those present each paid 18 Ft more than the planned contribution. Soon, however, one of the absentees arrived, so they recalculated the cost per person present. As a result, each of the first contributors received 10 Ft back. How many people were in the group, and how much did the gift item cost?
I. solution. Let's call the later arriving companion Anna, and the other distant one Bözsike. After the first arrangement of expenses, both Anna and Bözsike owed 18:2=9 Ft to each present person. With the returned 10 Ft, Anna settled her own debts, and with the additional 1 Ft, she reduced Bözsike's debts. With the second distribution of expenses, Anna also became a creditor to Bözsike, by 8 Ft, which is the amount more she paid than planned. This way, Bözsike's total debt did not change, so the sum of the 1 Ft reductions is 8 Ft. Thus, in the first payment, 8 companions participated, and there were 10 of them in total. Furthermore, Anna paid 8 * 10 Ft, so 72 Ft fell on one share, and the gift cost 720 Ft.

Vesztergombi György (Budapest, Piarist g. III. grade)

II. solution. We solve the problem in the following rephrased form, which does not change the numerical relationships.

"A group of friends all had the same amount of money. One of them left and divided their money equally, giving 8 Ft to each remaining person. Another one also left, and they also divided their money equally. This way, everyone ended up with 18 Ft more than they originally had. How many were there, and how much money did they have in total?"

Suppose the second one also initially gave everyone only 8 Ft. Thus, they had 16 Ft left: what they would have given to the first one who left, and what they received from them. From this 16 Ft, they could increase each remaining person's share by 2 Ft, so 8 remained, and initially, there were 10. The first one gave to 9 companions, so they had 72 Ft, and the group had 720 Ft in total.

Kóta József (Tatabánya, Árpád g. III. grade)

Except for 4, all solutions provided a mechanical answer based on a system of equations. We present only one of these.

III. solution. If there were $x$ people, and each person was to pay $y$ Ft, then first $x-2$ people paid $y+18$ Ft each, and then there were $x-1$ people, and each contributed $y+8$ Ft to the gift. Thus, the cost of the gift is once $x y$, once $(x-2)(y+18)=x y+18 x-2 y-36$, and once $(x-1)(y+8)=x y+8 x-y-8$. We get equal values if we subtract $x y$ from each:

$$
18 x-2 y-36=8 x-y-8=0
$$

From the last part of the equality,

$$
y=8 x-8
$$

Substituting this into the first part,

$$
18 x-16 x+16-36=2 x-20=0
$$

So originally, there were $x=10$ people, each paying $y=8 \cdot 10-8=72$ Ft, and the cost of the gift was $x y=720$ Ft.

Vincze Éva (Budapest, Kossuth Zsuzsa g. II. grade)