[ Processes and Operations ]

Between neighboring camps, it takes 1 day to travel. The expedition needs to transfer 1 can of food to the camp located 5 days away from the base camp and return. At the same time:

- each member of the expedition can carry no more than 3 cans of food;
- in 1 day, he consumes 1 can of food;
- food can only be left in camps.

What is the minimum number of cans of food that will need to be taken from the base camp for this purpose?
Suppose that each camp has a trading tent where delivered canned goods are sold. Let the price of one can in the base camp be one ruble, and in each subsequent camp, it is three times greater than in the previous one. In this case, the price of cans delivered to any camp is no less than the price of cans taken from the base camp. Indeed, to carry a can from some camp to the next, we need to take at least three cans, two of which will be consumed on the way there and back, respectively. The cost of a can in the fifth camp will be $3^{5}=243$ rubles, which means that no fewer than 243 cans need to be taken from the base camp.

Now let's prove that 243 cans will suffice. For this, 81 expedition members set out from the base camp. 54 of them, carrying one can each to the first camp, immediately return. The remaining 27 people, taking three cans each, proceed to the second camp. 27 cans are left in the first camp to ensure their return from the first camp to the base camp. This happens in each camp. Two-thirds of the arriving participants return, while one-third proceeds further. Thus, one person will set out from the fourth camp to the fifth. He will deliver the coveted can to the fifth camp.