Esmeralda arrives at a river bank bringing with her a goat, a wolf, and a head of cabbage. The only available means of transportation is a little boat which will hold Esmeralda and no more than one of her possessions. The wolf will eat the goat if they are left alone together, but won't touch the cabbage. The goat, however, will eat the cabbage if it gets a chance. How can Esmeralda cross the river with all her possessions intact?

1. bring the goat over and go back mt.
2. bring the wolf over and bring back the goat
3. bring the cabbage over and go back mt
4. bring the goat over.

I'm just having a hard time translating that to 3 men with 3 different amounts of money since there are 6 entities vs just 3: 3 men + 3 suitcases vs 2 animals and a head of cabbage

Now you should be able to solve the puzzle I posed.>

Ah, the one before that triviality (goat, cabbage,wolf). Sorry, did not see it before. It as well does not pose any problems:

1. Dick takes his 500 to the other side and returns mt.

2. Harry takes Tom's 300 to the other side and returns mt.

3. Tom and Dick cross the river and Tom returns with his 300.

4. Harry take his 800 to the other side and Dick returns with his 500.

5. Ton and Dick cross the river and Harry returns mt.

6. Harry takes the 300 to the other side and Dick returns mt.

1. Why is (-1)*(-1)=1 in the rational numbers? (minus times minus gives plus)

2. Why is 1*v=v an approriate axiom in the definition of a vectorspace?

But you didn't meet the conditions of the puzzle, ...>

Not really, as 500+300=800 and the rules say:

< al wazir: ... None of the men trusts the others. They will not allow any individual to remain alone on either bank with more than his share of the loot, or to take more than his share in the boat. ...>

Here's one to keep things going:

My second thought was, keep the shoes on. The trick must be to carry gold in *both* directions; and maybe some trips will be with *someone else's* bars.

Suppose that two men are already across with all their gold. Then the third one can make two trips, carrying one bar on one trip and two on the other, in order to transport his gold (or someone else's share, if it's less than his own) while staying within the rules.

The solution might begin with any of the three making the requisite two trips to carry his entire share across. So that takes care of the first man and the last. Two out of three ain't bad, as the saying goes, but it's not a solution. How can we get that middle guy across?

If at any point one man is on one side by himself, there can be no more than three bars of gold there, because any subset of four adds up to more than 24 ounces, which exceeds the largest share (Harry's).

That first man across can return after he's transported three bars, but then the next man can't bring any gold, because if he did he would be alone with four or more bars. So the first man must stay where he is. The next move must be for a second man to cross, carrying one or two bars of gold. But that leaves the third man all alone on the other side with four or five bars of gold, which is more than his share.

Alternatively, the first man could bring fewer than three bars, then return. But the next man across is limited in how many he can bring, because the total at that end must not exceed three, and we're back to the same situation as when only one man is doing the carrying.