2
$\begingroup$

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?

$\endgroup$
4
  • 5
    $\begingroup$ The question is a duplicate: Travellers across a desert $\endgroup$ – Jaap Scherphuis Oct 30 '20 at 22:28
  • 2
    $\begingroup$ Actually, it is not quite a duplicate, since the other question asks for at least two people to reach the other camp. It seems that this question does not allow that. Nevertheless, the fact that the distances match exactly indicates that the questions are closely related. $\endgroup$ – Jaap Scherphuis Oct 30 '20 at 22:42
  • $\begingroup$ @JaapScherphuis I also agree that this is not a duplicate question. It makes me sad that as it stands my question has been flagged this way. Can it be reopened?. $\endgroup$ – Chris Steinbeck Bell Oct 31 '20 at 23:06
  • $\begingroup$ It's successfully reopened now. $\endgroup$ – Rand al'Thor Nov 1 '20 at 13:18
1
$\begingroup$

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 $a=b=c=1$.

$\endgroup$
6
  • $\begingroup$ I got the same tidy answer by wondering how many helpers are required to complete the trip in the minimum 8 days. Now i'm having fun wondering about a neat graphical way to find the fewest days to cover the distance with no helpers. $\endgroup$ – humn Oct 31 '20 at 0:53
  • $\begingroup$ @JaapScherphuis The source of confusion here into trying to solve the problem by my own was that helpers seem to carry max 5 days of supplies in their backbacks. So when you depart you have to get a huge amount of supplies not only for yourself but also for the helpers to make you get one place and for them to safely return, is this the right interpretation?. $\endgroup$ – Chris Steinbeck Bell Nov 8 '20 at 2:46
  • $\begingroup$ @JaapScherphuis I'm stuck on why do the explorer ends with $2+3a+b-c$ after day $3$?. Why does $a$ gets tripled and $c$ is negative?. Can you please explain this part?. I don't know where $4+3a+4b+4c$ and $3+3a+b+3c$ comes from? what are these?. This is where I'm stuck, can you please do a clarification?. Using your inequalities I get to $2b\geq 1+c$ I guess that you intend to say that from here you landed to $b\geq 1$ and $c\geq 1$ is this right?. $\endgroup$ – Chris Steinbeck Bell Nov 8 '20 at 3:18
  • $\begingroup$ @JaapScherphuis The last part which I'm getting is that the sum of from solving the equation would be after getting to know $a,b,c$ then $1+1+1=3$, hence three helpers?. I would really appreciate if you could answer these doubts because I'm very confused. $\endgroup$ – Chris Steinbeck Bell Nov 8 '20 at 3:21
  • $\begingroup$ @ChrisSteinbeckBell You can write down exactly what happens each day just as I did under the first spoiler block. So during the first day the $1+a+b+c$ people use up $1+a+b+c$ rations, they carried $5(1+a+b+c)$, leaving $5(1+a+b+c)-(1+a+b+c)=4+4a+4b+4c$ by the end of the first day. The $a$ people that return need to take $a$ rations, leaving $4+3a+4b+4c$ for the rest of the party to use from day 2 onwards. The $1+b+c$ remaining people need to be able to carry this, so $5(1+b+c)\ge4+3a+4b+4c$. $\endgroup$ – Jaap Scherphuis Nov 8 '20 at 8:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.