This is a problem I once found in a math textbook a tutor used with me.
A group of explorers is trying to get from their base camp to another camp which is eight days' walk through the desert.
However, each explorer can only carry five days' worth of supplies. How many explorers need to venture out to make sure that at least two of the explorers make it to the second camp and the rest make it safely back to base camp before being stranded without supplies?
This problem is somewhat similar to the camel transporting bananas puzzle, but this time there are multiple people who can exchange supplies between each other.
Answer
Six explorers, according to the following schedule:
Day 1: 6 explorers start with 28 rations and end with 22. Two return with 2 rations (1 per).
Day 2: 4 explorers start with 20 rations and end with 16. One returns with 2 rations.
Day 3: 3 explorers start with 14 rations and end with 11. One returns with 3 rations.
Day 4: 2 explorers start with 8 rations and end with 6.
Day 5: 2 explorers start with 6 rations and end with 4.
Day 6: 2 explorers start with 4 rations and end with 2.
Day 7: 2 explorers start with 2 rations and end with 0.
Could "almost" do this with 5 and just let them get a little hungry but the six is needed just for one meal. Note that there's no need to load everyone to max at the beginning since 4 explorers can only carry 20 meals.
UPDATE
As pointed out in comments, the solution above assumes that they eat at the end of the day (where day 8 should be "start with 0 and arrive hungry"). Below follows the rather similar solution for the more reasonable scenario of eat at the start or during the day.
Day 1: 8 explorers start with 40 rations and end with 32. Two return with 2 rations (1 per).
Day 2: 6 explorers start with 30 rations and end with 24. Two return with 4 rations (2 per).
Day 3: 4 explorers start with 20 rations and end with 16. Two return with 6 rations (3 per).
Day 4: 2 explorers start with 10 rations and end with 8.
Day 5: 2 explorers start with 8 rations and end with 6.
Day 6: 2 explorers start with 6 rations and end with 4.
Day 7: 2 explorers start with 4 rations and end with 2.
Day 8: 2 explorers start with 2 rations and end with 0.
This solution is more symmetrical, which feels better somehow.
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