Originally posted by HandyAndy 79 is the correct answer. Good work!
Thanks. Got a little fed up with the coconut to the monkey each time!
To make it a Diophantine puzzle, rather than give the answer '7', shouldn't you leave it open with 'each sailor ended up less than 10' or something like.
I don't think it works for any other number 1-10, but if you HAD posed it that way, I wouldn't have had a chance!
Originally posted by Blood On The Tracks Thanks. Got a little fed up with the coconut to the monkey each time!
To make it a Diophantine puzzle, rather than give the answer '7', shouldn't you leave it open with 'each sailor ended up less than 10' or something like.
I don't think it works for any other number 1-10, but if you HAD posed it that way, I wouldn't have had a chance!
Why not start with each getting one plus the monkey's one, and see when/if it leads to partial coconuts. Then start with 2. I haven't bothered doing this.
Is there any starting share that never leads to partial coconuts? This would not be provable by brute force, would it? It is an unending task, no?
Originally posted by Blood On The Tracks Thanks. Got a little fed up with the coconut to the monkey each time!
To make it a Diophantine puzzle, rather than give the answer '7', shouldn't you leave it open with 'each sailor ended up less than 10' or something like.
I don't think it works for any other number 1-10, but if you HAD posed it that way, I wouldn't have had a chance!
The posted puzzle is one of several versions available on the Internet. For non-specialists it's hard enough.