Problem 3.8. A small raft was swaying by the riverbank. Five mice weighing 70 g each, three moles weighing 90 g each, and four hamsters weighing 120 g each approached the shore. What is the minimum number of grams the raft must be able to support so that all the animals can cross to the other side, possibly in several trips "back and forth"? The raft cannot move across the river without a rower.
Answer: 140.

Solution. If the raft can carry less than two animals, then they will not all be able to cross. Indeed, in this case, one animal would have to row back and forth on the raft, without the possibility of staying on the other side of the river (since the raft cannot return without a rower).

Therefore, the raft must be able to carry at least the two lightest animals, i.e., at least $70 \cdot 2=140$ grams. We will show that if the raft's capacity is 140 grams, then all the animals can cross to the other side of the river.

We will designate two mice as "rowers," and call them $A$ and $B$. The two mouse rowers row to the other side; $B$ stays there, while $A$ returns with the raft. Then, one of the other animals rows to the other side. Mouse $B$ returns the raft.

Thus, one animal has been ferried, and the rest of the animals and the raft are back in the initial position. Therefore, by repeating the same operations, we can ferry the next animal. And so on, until all the animals are ferried.

At the very end, $A$ and $B$ will row to the other side, and all the animals will have crossed.

## 4th grade