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Two thieves, Bo and Jo, steal 7 gold bars and are driving away when their car is wrecked in an accident. They-and the loot- is intact.

They are at Point A at 8 PM and must go to Point B which is 1 mile away. There is a small train station at B. The last train will leave at 11 PM. Before that, there are trains leaving at 9 and 10 PM also. Realistically, they need to be there 5 minutes early to load the loot.

Both can carry only 1 bar at a time. They cannot use any other means or methods to carry the bar.

Bo can carry the bar at 2 mph and Jo can carry the bar at 1.5 mph. They both can go 3 mph without the bar.

Here is the problem. Both are smart but greedy and absolutely do not trust each other. Either one of them will run away if he has a chance and (of course) 4 or more bars to take with him. But he must do that safely with no chance for the other guy to catch him. And he can only take off on the train. No other way. So if one of them is at B with 4 bars, say around 9:50 PM and the other one is more than 10 minutes away, then he will take off on the 10 PM train. Just an example.

They are greedy and do not want to leave any bars behind.

What strategy must they come up with so they can take all 7 bars with them on the last train and no one has a chance to run away with 4 or more bars?

Linked to The Thieves and the Gold Bars

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    $\begingroup$ BoJo is stealing gold => we are in United Kingdom => you cannot trust the trains to be on time. This problem is very tough. $\endgroup$
    – Evargalo
    Sep 28 at 12:37
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Assuming that Bo and Jo cannot set up secret stashes from each other (as otherwise the problem is fairly trivial), but assuming that they can set up a shared stash, one strategy they could use is to:

Set up a cache anywhere between 18 ∕ 25 mile and 13 ∕ 16 mile from A, where they will first deliver all the bars to. The cache will be far enough from B to prevent either of them stealing bars and absconding on the 10:00 train. Once they've brought all bars to the cache, they can take the bars to B and catch the 11:00 train.

Proof:

It takes Bo 30 minutes to transport a bar 1 mile, and Jo 40 minutes. It takes them 20 minutes to walk 1 mile unladen. A round trip takes Bo 50 minutes, and Jo 60 minutes. It thus takes Bo 50 + 30 = 80 minutes to transport 2 bars from A to B, and Jo 60 + 40 = 100 minutes, so neither of them is stealing onto the 9:00 train.

If they take bars straight from A to B, Jo will arrive with the 4th bar (his 2nd) after 100 minutes, at 9:40. At this point Bo has completed two round trips and is back at A - even if he goes back to B unladen he won't arrive until 20 minutes later, i.e. 10:00, by which point Jo is on the departing train with 4 bars of gold. So clearly they must leave bars at an intermediate location.

Let the position of the intermediate cache be x miles from A. At maximum efficiency, the 4th bar will be delivered by Jo after 100x minutes. Assume Bo has hidden in wait the whole time, and will immediately start stealing the bars as soon as Jo sets back off for A.

Bo has to transport 4 bars, which takes him (50 × 3) + 30 = 180 minutes per mile. As he has to transport them (1 − x) miles, this takes 180 − 180x minutes.

The total elapsed time is thus 180 − 80x minutes, which we thus require to be more than 115 to prevent Bo getting onto the 10:00 train. 180 − 80x > 115, so 80x < 65 and x < 13 ∕ 16.

Note if Jo wants to steal the bars, he also has to start his theft as soon as he drops off his 2nd bar, which is the same time Bo starts his attempt - but as Jo is slower than Bo, if Bo can't steal the bars, neither can Jo.

Now assured of co-operation, we just need to sort out the logistics. Bo, being faster, will deliver 4 bars to the cache, and Jo 3.

Bo completes his task after ((50 × 3) + 30)x = 180x minutes.

Jo completes his task after ((60 × 2) + 40)x = 160x minutes.

If we have Bo then take 4 bars from the cache to B, it will take him a further 180(1 − x) minutes, which means he will only finish after a total elapsed 180 minutes - i.e. at 11:00, too late to load the train. Jo must thus use his headstart to help - Jo will pick up 4 bars from the cache, and Bo 3.

Transporting 3 bars takes Bo (50 × 2) + 30 = 130 minutes per mile, so transporting 3 bars for (1 − x) miles takes him 130 − 130x minutes for a total of 130 + 50x minutes.

Transporting 4 bars takes Jo (60 × 3) + 40 = 220 minutes per mile, so transporting 4 bars for (1 − x) miles takes him 220 − 220x minutes for a total of 220 − 60x minutes.

Bo will finish first if 130 + 50x < 220 − 60x i.e. if 110x < 90. But since x < 13 ∕ 16, 110x < 1430 ∕ 16 = 89.375, so this is always true.

When Bo finishes, Jo will be (220 − 60x) − (130 + 50x) = 90 − 110x minutes away from B. We could have Jo carry the last bar the remaining distance, but it will be quicker to have Bo carry it. In fact, if x < 3 ∕ 4, we must have Bo pick up the bar, since 220 − 60x > 220 - 45 = 175 in this case.

Bo will thus rendezvous with Jo at a combined approach speed of 4.5mph, which will take (90 − 110x) × 1.5 ∕ 4.5 = 30 − (110 ∕ 3)x minutes. Total time elapsed is now 160 + (40 ∕ 3)x minutes.

The remaining distance takes 60 − (220 ∕ 3)x minutes at 1.5mph, so at 2mph it will take Bo (60 − (220 ∕ 3)x) × 1.5 ∕ 2 = 45 − (165 ∕ 3)x minutes. Jo is faster and walks ahead to B. Total time elapsed is 205 − (125 ∕ 3)x minutes.

We thus require 205 − (125 ∕ 3)x < 175, in order for Bo to arrive with the final bar in time to load the train. Thus (125 ∕ 3)x > 30, and so x > 90 ∕ 125 = 18 ∕ 25.

Thus, so as long as the cache is between 18 ∕ 25 and 13 ∕ 16 miles from A, Bo and Jo can get all the bars to the train in time. The process is quickest with the cache at just shy of 13 ∕ 16 mile (since exactly 13 ∕ 16 mile lets Bo steal the bars), taking them just over 171 minutes.


Addendum:

It's possible to optimise this further. My above solution has Bo directly pick up a bar from Jo, but this is inefficient, since Jo is now walking towards B unladen. To make best use of their time, when walking towards B they should always be carrying a bar, and never carrying a bar away from B. This is possible by dropping bars partway along to be picked up later.

The optimal solution has both Bo and Jo finishing at the train station at the exact same time. Otherwise, the one who finishes first could be taking some work off his colleague's hands.

Let b be the distance Bo carries bars for, and j be the distance Jo carries bars for. b + j = 7 miles.

The time taken by Bo is thus (50 × (b − 1)) + 30 = 50b − 20.

The time taken by Jo is thus (60 × (j − 1)) + 40 = 60j − 20.

Thus 50b − 20 = 60j − 20, therefore 5b = 6j. b = 7 × 6 ∕ 11 = 42 ∕ 11, j = 7 × 5 ∕ 11 = 35 ∕ 11.

This can work out as Bo and Jo each carrying 3 bars the whole way, with the 7th bar being carried by Bo 9 ∕ 11 of the way, and by Jo 2 ∕ 11 of the way.

As it turns out, 2 ∕ 11 < 1 - (13 ∕ 16); our cache is always at least 2 ∕ 11 miles from B. So Jo can transport one bar for 2 ∕ 11 miles, then drop it for Bo to take the rest of the way. It thus takes him 2 ∕ 11 × 60 = 120 ∕ 11 minutes to drop off that one bar, and ((60 × 2) + 40)(1 − x) = 160(1 − x) to drop off the other 3 at the station. His total time elapsed is 160x + 160(1 − x) + 120 ∕ 11 = 160 + 120 ∕ 11 = 170 and 10 ∕ 11 minutes.

Bo transports his 3 bars in 130(1 − x) as before. He goes back (1 − x) − 2 ∕ 11 = 9 ∕ 11 − x miles to pick up the bar Jo dropped, and bring it to the station. This takes 50 × (9 ∕ 11 − x) = 450 ∕ 11 − 50x minutes. His total time elapsed is 180x + 130(1 − x) + 450 ∕ 11 − 50x = 130 + 450 ∕ 11 = 170 and 10 ∕ 11 minutes, the same as Jo.

So, Bo and Jo both arrive just before 10:51, giving enough time to load the bars and get a coffee and a ham sandwich. This is actually independent of where they place the cache too, as long as it's within the limits derived earlier. The cache just stops either one absconding onto the 10:00 train.

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    $\begingroup$ Superb analysis. Hope you liked the puzzle. $\endgroup$
    – DrD
    Sep 28 at 20:25
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Bo and Jo must:

Pass the bars to each other and bar 6 must be dropped at a predetermined meeting point.

Bo's travel time is:

30 minutes with the bar; 20 without.

Jo's travel time is:

40 minutes with the bar; 20 without.


Bars 1 and 2 are the easiest to solve:

Bo and Jo leave at the same time. Bo drops bar 1 at 8:30 and Jo drops bar 2 at 8:40.

From here things get interesting, but essentially:

Bo must pick up bars and run them to Jo who drops them off at the train station.

Which means we must calculate:

The time it takes them to close the gap with each run. This can be expressed as: $$\frac{d}{b + j}$$ Where $b$ is the speed of Bo and $j$ is the speed of Jo.

Bars 3 and 4:

Bo arrives back at the start at 8:50 and starts running bar 3 to Joe. At this point, Jo has already been running back to Bo for 10 minutes and is half way back to the starting point. As such the pair are currently half a mile apart.

They can close the gap between them in 6 minutes: $$\frac{0.5mi}{2mph + 3mph} \cdot 60 = \frac{0.5}{5} \cdot 60 = 0.1 \cdot 60 = 6$$ As such, Bo passes the bar off to Jo at 8:56, at a tenth of a mile away from the starting point, then heads back.

Bo arrives back at the starting point at 8:58 covering a distance of a tenth mile in 2 minutes at a rate of 3mph. Jo arrives at the train station at 9:32 covering $\frac{9}{10}$ of a mile in 36 minutes at a rate of 1.5mph.

Since it took Jo 36 minutes to run bar 3 and it took Bo 2 minutes to get back to the starting point, Bo has 34 minutes to get bar 4 to Jo. However, it takes Bo 30 minutes to run one bar the entire mile. As such, bar 4 is actually delivered at 9:28, 4 minutes before bar 3.

Bar 5:

Bo departs the train station at 9:28 and Jo departs at 9:32. Since both can run at 3mph without a bar, Bo remains 4 minutes ahead of Jo, arriving at the starting point at 9:48.

At this point, Bo and Jo are currently a fifth of a mile apart and Bo runs bar 5 to Jo. $$\frac{0.2}{5} \cdot 60 = 0.04 \cdot 60 = 2.4$$ Bo passes bar 5 to Jo at approximately 9:50 (add 24 seconds). Bo then sprints back to the starting point, arriving at approximately 9:53 (subtract 12 seconds). Jo arrives at the train station at 10:22 covering $\frac{4}{5}$ of a mile in 32 minutes at a rate of 1.5mph.

Bar 6:

After some pre-planning, Bo and Jo determine that bar 6 needs to be dropped at just beyond the half way point; $\frac{3}{5}$ of the way to the train station.

Bo departs the starting point at 9:53 and runs bar 6 to the meeting point at a rate of 2mph, dropping the bar 18 minutes later at 10:11. Bo then runs back to the starting point, arriving at 10:23.

Jo departs the train station at 10:22 and picks up bar 6 from the meeting point at 10:30, covering 0.4mi in 8 minutes at a rate of 3mph. It takes Jo 16 minutes to run bar 6 from the half way point to the train station, so Jo arrives at the train station at 10:46 leaving 14 minutes until the 11pm train departs.

Bar 7:

Bo departs the starting point at 10:23 and sprints to the train station at a rate of 2mph, covering the entire mile in 30 minutes, arriving at 10:53. Just enough time to catch their breath and help Jo load the train.

Summary of Train Station Arrivals

Bar 1: 8:30
Bar 2: 8:40
Bar 3: 9:32
Bar 4: 9:28
Bar 5: 10:22
Bar 6: 10:46
Bar 7: 10:53

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    $\begingroup$ I don't think this works out. At the Bar 5 step, Jo should just stay at the train station with the 4 bars already there. Bo arrives back at the start at 9:48, and doesn't realize he's been double-crossed until 9:50. Jo hops on the 10 PM train with his 4 gold bars, as Bo can't get there until 10:08, which is too late to stop him. You'd need Jo and Bo to leave together to ensure they both actually go, but that's 4 minutes you don't have to spare. $\endgroup$ Sep 28 at 17:10
  • $\begingroup$ @NuclearHoagie nooooooooooo now I have to change my answer lol $\endgroup$ Sep 28 at 17:19
  • $\begingroup$ I'm just gonna leave it the way it is though. There's already a better answer and this was a valiant attempt lol $\endgroup$ Sep 28 at 17:20
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    $\begingroup$ Agree, it's a good effort to a tricky problem! It's tough with antagonistic agents who try to deviate from "the plan" at any opportunity. $\endgroup$ Sep 28 at 17:23
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Missed the 5 minutes early being important. T. Linnell has it done properly.

Switching from rates

Bo takes 30 min with the gold, 20 min back. Bo can only move 4 bars just barely in the 3 hours.

Jo takes 40 min with the gold, 20 min back. Jo can only move 3 in the 3 hours.

Solution?

If they hide their bars from each other in the station it should all be golden. They will always be 10 minutes apart so they can't watch each other hide bars.

Bigger question, how they get the bars on the train quickly without someone pushing the other out at the last second?

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    $\begingroup$ I don't know if the question was changed, but you are not accounting for "be at the station 5 minutes before the train leaves to load everything" $\endgroup$ Sep 28 at 14:47
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    $\begingroup$ Bo's 4th bar will there at 11 not 5 minutes before. $\endgroup$
    – DrD
    Sep 28 at 20:21

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