Barabás's grandmother has been setting aside some loose change in her purse for her grandson every lottery draw recently. The grandmother is an extremely precise lady, and she always follows these rules:

1) She only sets aside coins.
2) For each number drawn, she puts the corresponding amount of money into the purse, making sure to use the fewest number of coins.
3) After the draw, she always writes down how many of each coin she put into the purse.

After one draw, where 7 numbers were drawn from the first 35 positive integers, the grandmother noted that she had put $3 \mathrm{db} 20$, $6 \mathrm{db} 10$, $5 \mathrm{db} 5$, $9 \mathrm{db} 2$ and $3 \mathrm{db} 1$ forint coins into the purse. What were the numbers drawn?
A grandmother could only put one coin of 20, 10, 5, and 1 forints into the piggy bank at a time, since $2 \cdot 20 = 40 > 35$, and in the case of other denominations, two identical coins could be replaced by one. She could also put in at most two 2 forint coins at a time, since $2+2+2=5+1$ can already be represented by two coins instead of three. Thus, considering that the total value of the coins could not exceed 35 - at most four coins could be put into the piggy bank after each draw. Since the total number of coins put into the piggy bank is $3+6+5+9+3=26$ and $4 \cdot 4 + 3 \cdot 3 = 25 < 26$, in at least five draws, four coins were put into the piggy bank.

There are six four-term sums that do not exceed 35 and meet the requirements of the problem; these are:

$$
\begin{array}{rrr}
20+10+2+2=34 ; & 20+10+2+1=33 ; & 20+5+2+2=29 \\
20+5+2+1=28 ; & 10+5+2+2=19 ; & 10+5+2+1=18
\end{array}
$$

The number of 10 forint coins set aside is 6, and four of these four-term sums include a 10; therefore, these four sums, 34, 33, 19, and 18, are definitely among the drawn numbers. The total number of 20 forint coins is 3, so of the two four-term sums that include a 20 but not a 10, at least (or exactly) one is definitely missing.

Suppose the missing sum is $29=20+5+2+2$. By adding up the terms of the remaining five sums, we see that they include 3 20s, 4 10s, 3 5s, 7 2s, and 3 1s, leaving 2 10s, 2 5s, and 2 2s. However, from these, two different three-term sums that meet the problem's requirements cannot be formed. Therefore, 29 is also among the drawn numbers, and 28 is not, leaving 2 10s, 2 5s, and 1 each of 1s and 2s.

The 10s and 5s must appear in both three-term sums, with the 2 in one and the 1 in the other, so the two additional drawn numbers are 16 and 17.

The seven drawn numbers are thus: $34, 33, 29, 19, 18, 17, 16$.

Balázs Poronyi (Pécs, Janus Pannonius Gymnasium, 9th grade)