1. Honza, Jirka, Martin, and Petr organized a charity collection on the square. Soon, five passersby stopped by them one after another. The first one gave 3 Kč to Honza, 2 Kč to Jirka, 1 Kč to Martin, and nothing to Petr. The second one gave 8 Kč to one of the boys and nothing to the other three. The third one gave 2 Kč to two of the boys and nothing to the other two. The fourth one gave 4 Kč to two of the boys and nothing to the other two. The fifth one gave 8 Kč to two of the boys and nothing to the other two. Afterward, the boys realized that each of them had collected a different amount, and these amounts were four consecutive natural numbers. Which of the boys collected the least and which one collected the most money?
SOLUTION. Altogether, the boys received $3+2+1+8+2 \cdot 2+2 \cdot 4+2 \cdot 8=42$ Kč. This number can be uniquely expressed as the sum of four consecutive natural numbers: $42=9+10+11+12$. Thus, the four boys (in some order) picked the amounts of $9,10,11$ and 12 Kč.

No boy could receive 8 Kč from both the second and the fifth passerby (otherwise he would have at least 16 Kč, but the most any of the boys could have was 12 Kč). So, from the second and fifth, three boys each received 8 Kč and one of them received nothing. At most one of these three boys could have received 4 Kč from the fourth passerby, otherwise at least two boys would already have at least 12 Kč. Therefore, the fourth passerby must have given 4 Kč to exactly one of them and 4 Kč to the remaining boy. Without the money from the first and third passersby, the boys thus have $12,8,8$ and 4 Kč. The boy who received a total of twelve crowns from the second, fourth, and fifth passersby could not have received anything from the first and third passersby, as he would then have more than twelve crowns. The one who received a total of $4 \mathrm{Kč}$ from the second, fourth, and fifth passersby must have received the maximum possible amount from the first and third, i.e., $3+2=5$ Kč, otherwise he would have less than 9 Kč in total (he thus received exactly 9 Kč and has the least). So, the one who received 3 Kč from the first passerby is Honza, and the one who received nothing from the first is Petr.

The reasoning can easily be completed and we can show that the described distribution is indeed possible. As we already know, Honza picked 9 Kč and Petr 12 Kč. Jirka, who received 2 Kč from the first, could not have received anything from the third, so he received a total of 10 Kč, and Martin 11 Kč. All the reasoning can be neatly organized into a table, which we fill in step by step:

| 1 | 2 | 3 | 4 | 5 | $\Sigma$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
|  | 8 |  | 0 | 0 |  |
|  | 0 |  | 0 | 8 |  |
| 0 | 0 | 0 | 4 | 8 | $12 \longrightarrow \mathrm{P}$ |
| 3 | 0 | 2 | 4 | 0 | $\leqq 9 \longrightarrow \mathrm{H}$ |
| $\mathbf{1}+\mathbf{2}+\mathbf{3}$ | $1 \times 8$ | $2 \times 2$ | $2 \times 4$ | $2 \times 8$ |  |

GUIDING PROBLEMS: