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I have a puzzle stated as follows: There are three friends travelling to a party from the same location. The party is $320$ kilometres away, and they must arrive at the same time. They have two motorcycles to use - one that can go $60 \text{km/h}$, and one that can do $80 \text{km/h}$. They can also run at $40 \text{km/h}$. Each motorcycle can only carry one person. These friends can swap modes of transport at the end of each hour. They are not allowed to wait at a spot for the rest to catch up, but they can leave either of the bikes for others to pick up. How long does it take them?

Not a very realistic problem, but it's challenging me. I tried setting up a system of equations and was hoping to solve with matrices but I ended up with too many unknowns. I also tried "brute force" thinking it, but I can't make it such that they only swap modes every hour. What's the best way to go about this? Thanks for your time.

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    $\begingroup$ They can run 40km in an hour? Sign them up for the Olympic marathon competition! $\endgroup$
    – Gerry Myerson
    Commented Aug 24, 2020 at 12:51
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    $\begingroup$ I've also seen the problem stated as ships travelling 40, 60 and 80 light years per hour, so realism is not a theme with this particular puzzle it seems. $\endgroup$
    – James Oswald
    Commented Aug 24, 2020 at 13:00
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    $\begingroup$ Should they arrive with both transport at the end of the travel? $\endgroup$
    – Tortar
    Commented Aug 24, 2020 at 13:03
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    $\begingroup$ It is not clear what you mean that they can change modes every hour. If one person rides the fast bike for a while and leaves it for somebody else to pick up, does each change have to be on the hour, or is it that one person cannot change more than once per hour? $\endgroup$
    – Ross Millikan
    Commented Aug 24, 2020 at 13:04
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    $\begingroup$ One rides the fast motorbike for $2$ hours and runs for $4.$ Another runs for $4$ hours, picks up the fast motorbike, and rides for $2$. The third rides the slow motorbike for $4$ hours, abandons it, and walks for $2$. All arrive in $6$ hours, without the slow motorbike. Is this a legitimate solution? $\endgroup$
    – saulspatz
    Commented Aug 24, 2020 at 13:13

1 Answer 1

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For $i=1,2,3$, let us say the participant $i$ rides the fast bike for $f_i$ hours, the slow bike for $s_i$ hours and runs for $r_i$ hours. By assumption, $f_i,s_i,r_i$ are all non-negative integers. Also, by symmetry, and to eliminate trivial solutions, we can assume $f_1>0, s_2>0$, that is, participant $1$ starts out on the fast bike and participant $2$ starts out on the slow bike.

We have $$80f_i+60s_i+40r_i=320,\ i=1,2,3$$ or $$4f_i+3s_i+2r_i=16$$ and we observe that $s_i$ is even.

Now the fast motorbike cannot travel more than $320$ miles, and the slow motorbike cannot travel more than $240$, since it travels an even number of hours. That makes a total of $$3\cdot320-320-240=400$$ miles to be covered running. We have $4$ hours for each bike and $10$ hours running for the $3$ participants, so the trip cannot take less than $6$ hours.

Here is a six-hour solution. One rides the fast motorbike for $2$ hours and runs for $4$. Another runs for $4$ hours, picks up the fast motorbike, and rides for $2$. The third rides the slow motorbike for $4$ hours, abandons it, and walks for $2$. All arrive in $6$ hours, without the slow motorbike.

I had thought this was the only solution, but that is not true.

We have $$80f_i+60s_i+40r_i=320,\ i=1,2,3$$ or $$4f_i+3s_i+2r_i=16$$ and we observe that $s_i$ is even. We must have $f_i+s_i+r_i$ constant. With such small numbers, we can check for possible solutions by brute force. I wrote a little python script to do so, and found $5$ solutions:

(1, 0, 6) (0, 2, 5) (0, 2, 5)
(1, 0, 6) (0, 2, 5) (1, 0, 6)
(1, 2, 3) (1, 2, 3) (2, 0, 4)
(2, 0, 4) (0, 4, 2) (2, 0, 4)
(2, 0, 4) (1, 2, 3) (1, 2, 3)

Each triple is of the form $(f_i,s_i,r_i)$. The last three solutions each take $6$ hours. The fourth is the one I gave above. To see that the third actually leads to a solution of the problem, let participant $1$ ride the fast motorbike for an hour and run for an hour, traveling $120$ miles in $2$ hours. Let participant $2$ ride the slow motorbike for $2$ hours, also traveling $120$ miles in $2$ hours. Participant $3$ runs for $2$ hours, arriving where participant $1$ left the fast bike. Participant $1$ rides the slow motorbike for $2$ hours and then runs for $2$ hours, arriving at the goal. Participant $3$ rides the fast motorbike for $2$ hours and then runs for $2$ hours. Participant $2$ runs for $3$ hours, arriving at mile $240$, where participant $3$ left the fast motorbike, and rides it for $1$ hour.

I haven't checked the fifth solution, but I imagine it will work also. It's possible that there's more than one protocol

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  • $\begingroup$ Ah, the observation that $s_i$ is even almost immediately settles the optimality question. As for the third vs fifth solution, they are symmetric (swap first & third person). As for "protocol", even for the fourth solution the swapping of rides can be different from your canonical form, e.g. the two people using the fast bike can swap more than once and still stick to $(2,0,4)$ totals. ... I wonder what is the optimal solution without the integer constraint. That sounds like a more fun problem than the OP! $\endgroup$
    – antkam
    Commented Aug 25, 2020 at 19:58
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    $\begingroup$ @antkam, you might be interested in Chvatal's On the Bicycle Problem (1983) $\endgroup$
    – RobPratt
    Commented Aug 25, 2020 at 20:05
  • $\begingroup$ @RobPratt - THANK YOU! I had no idea variations of this problem can run into issues of real analysis and swapping rides infinitely often and limits and such. Neat! $\endgroup$
    – antkam
    Commented Aug 25, 2020 at 20:13

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