Last year I posed three mathematical puzzles in a blog. This time there’s just one, but don’t just read it and then look at the answer – it’s a really good puzzle to spend some time trying to solve; most people give up, and yet when you see the solution, it’s surprisingly straightforward. And please note: there are no catches, no tricks, no sneaky wordplay; the puzzle is exactly as I present it.
Four people are being chased by a dragon. They reach a bridge which, if they can all cross, they can then destroy, and this will stop the dragon in its tracks. But the following limitations must be observed.
- The bridge is quite weak, and can only take the weight of two people at any one time.
- One of the four has a twisted ankle and will take 10 minutes to cross the bridge. Another has blisters and takes 5 minutes. The two remaining people can run, one of them taking 2 minutes and the other taking 1 minute.
- It’s night time, and the bridge has no handrails, so it’s imperative that anyone crossing the bridge carries a torch. They only have one torch between them, so if two people cross together they must stay together (so for example, if the 2 minute and the 5 minute cross together, they will take 5 minutes).
- The bridge is too long to throw the torch back; it must always be carried.
- The dragon is 17 minutes behind – do they make it over the bridge safely?
Looking at the problem logically, it should be clear that every time two people cross, one must return with the torch. First, then, let’s calculate how many crossings they need to make.
2 over, 1 back = 1 safely over
Another 2 over, 1 back = 2 now safely over
Another 2 over – and all 4 have now made it
The person bringing the torch back should always be the fastest possible: the 1 minute (1 min) for preference, the 2 min if necessary. If at any stage you send the 5 min back, then this would mean the 5 min making three crossings altogether, using up 15 valuable minutes. And how should the pairs be made up? At first sight, it seems sensible to pair a fast with a slow – say 1 min with 10 min, and 2 min with 5 min. But again, that would already use 15 minutes. The key to the solution is to pair the 5 with the 10; but if you send them over first, one of them will have to come back.
So, the outcome of all that logic, is that we should send the 2 over with the 1, and then send the 1 back (3 minutes so far), and then send the 5 over with the 10 (13 minutes so far). What next? Well, the 2 is already over there, so send the 2 back (15 minutes so far), and then finally the 1 and the 2 both come over again (17 minutes). The part where most people get stuck is in assuming that the person coming back is always one of those who has just gone over. The schematic below shows how it works.