How to find the least amount of helpers required to complete a voyage?

Question

The puzzle is as follows:

Vincent is a scientist and is about to begin a journey in Antarctica, as he must reach his laboratory at the next station. However, the only buggy available at the base station is undergoing engine maintenance and cannot be used. For this reason he must go on foot. According to his map the trip will take 8 days. However, the weight of supplies that he can carry in his backpack will only last 5 days. For this reason, he needs to recruit one or more assistants since they can carry a weight for 5 days of supplies. The supplies can be shared between Vincent or his assistant. For example, if only the helper's supplies are consumed between him and Vincent on a day of the expedition. The assistants' supplies will last for only 3 days. Since it is required that the assistants must return safely to the base camp without running out of supplies, what is the least number of assistants which Vincent needs to recruit?

The alternatives in my book are as follows:

1. 5 assistants
2. 6 assistants
3. 3 assistants
4. 2 assistants

This problem is from my Reason and Logic book of puzzles and seems to be an adaptation from a reprinted version of Martin Gardner's 70s book of Puzzle Carnival.

This problem has left me go in circles too many times. Can someone help me?

I attempted to sketch the situation.

Assuming Vincent is on point A (base camp) and his laboratory on point B:

$$\textrm{A---------------------B}$$

For this scenario:

Vincent would advance to a point in between $AB$ which I would call $C$:

$$\textrm{A-----------C----------B}$$

Since he cannot move forward because his supplies don't last enough to complete the expedition, the problem mentions that he must recruit some helpers to carry more supplies. So in the preceding sketch point C is Vincent plus a helper.

Since the helper can also carry supplies for 5 days, he can advance the same distance attained by Vincent but he cannot return without taking Vincent's supplies with him, therefore, if he only recruits one, it will be insufficient to fulfill his needs. So he must recruit more than one.

This is the part where it becomes troublesome:

Vincent would advance with his helpers to a short distance, they would share some supplies, sufficiently enough for him to keep advancing and to ensure that at least one helper return safely. But we will get to a point where the distance covered by Vincent and his helpers is wide enough to constrain the further advancing without compromising the helpers safely return. So this means that a set of helpers will stand by along the path to act as a relay network allowing a safely return.

At least this is what I concluded. But I don't know how to put this into a graphic or a sketch. Can someone help me with an answer analyzing this strategy? I mean using a visual aid to illustrate how this voyage is taken? I'm sorry if this is too much for a puzzle which may seem easy but for me, I find it some confusing because I'm getting tangled with paths. Can an answer include the requested feature as mentioned lines above? Overall, is my analysis or logic good?

Answer 1

The answer to the problem is as follows:

Day 1: 4 people start with 20 days of supplies, end with 16.
Then 1 person turns back, with 1 day of supplies.
This leaves 3 people with 15 days of supplies.

Day 2: 3 people start with 15 days of supplies, end with 12.
Then 1 person turns back, with 2 days of supplies.
This leaves 2 people with 10 days of supplies.

Day 3: 2 people start with 10 days of supplies, end with 8.
Then 1 person turns back, with 3 days of supplies.
This leaves 1 person with 5 days of supplies.

Days 4-8: 1 person travels 5 more days using 5 days of supplies.

So only 3 helpers are needed.

You can derive the solution as follows:

Suppose $a$ people turn back after day 1, $b$ after day 2, and $c$ after day 3. There is no need for the helpers to travel any further, because their aim is to put the explorer at a distance of 5 days from the other camp with 5 days of supplies.
Then $1+a+b+c$ people set off with $5(1+a+b+c)$ supplies. You can track their supplies after each day, and you'll find that the explorer ends up with $2+3a+b-c$ supplies after day 3, so we need $2+3a+b-c=5$.
The supplies at the end of day 1 and day 2 must be able to be carried by the remaining people, and that gives $5(1+b+c)\ge4+3a+4b+4c$ and $5(1+c)\ge3+3a+b+3c$. If you use the equation to eliminate $3a$ from the inequalities, you find that $b\ge1$ and $c\ge1$. The smallest values that satisfy the equation $2+3a+b-c=5$ is then $a=b=c=1$.