Yep, you all got it - it's too easy for people who know mathematics. bwtaylor's solution skipped a few dozen steps, so here's a more complete solution:
Each sailor gets n coconuts,
So before they split up the pile there were 5n+1
coconuts
Before the 5th sailor split up the pile there were
(5/4)(5n+1)+1
Before 4Th sailor (5/4)((5/4)(5n+1)+1)+1
Before 3 (5/4)((5/4)((5/4)(5n+1)+1)+1)+1
Before 2
(5/4)((5/4)((5/4)((5/4)(5n+1)+1)+1)+1)+1
Before 1
(5/4)((5/4)((5/4)((5/4)((5/4)(5n+1)+1)+1)+1)+1)+1
which is the original number of coconuts, N.
N = (5n+1)(5/4)^5 + (5/4)^4 + (5/4)^3 + (5/4)^2 + (5/4) +
1
= 15n + 11 + (n+1)/4 + (n+1)/128 + (n+1)/1024
Since 4 and 128 are both multiples of 1024 we can simply say
n = 1024c-1, c = 1,2,...
N = 15625c - 4, c = 1,2,...
The seven bridges of Koenigsberg
This one is a classic from topology. In the Prussian city of Koenigsberg (Modern ???) there are two islands in the river with seven bridges thus:
---v--v-------v----
/ | | | \
/ --^--^--- --^-- \
--- | | | | -------
| >--< |
--- | | | | -------
\ --v--v--- --v-- /
\ | | | /
---^--^-------^----
A popular passtime of the inhabitants was to try to cross
every bridge only once. Prove that it can not be done.
Ha! The thirteen coin problem!
bwtaylor:
It's possible to
do this with thirteen coins. Work that one out! 8)
Hmm... Well you'll be having trouble - It can't be done. I didn't count up how many coins were being measured in each go, obviously the scales are supposed to balance with each measure, but mine weren't. 12 is the maximum.
At most you can search for (((3^n)-1)/2)-((((3^n)-1)/2)%2) coins so that the scales will balance. Searching for ((3^n)-1)/2 coins will require an extra measuring to callibrate the scales if it is not divisible by two. The -1 is required to eliminate the zero permutation (where one coin is not measured) since it does not tell us whether the forgery is heavier or lighter
For those who are having trouble with 12 coins, the trick is to stop thinking like a programmer and start thinking like a mathematician. Trying to walk a tree in your head is going to lead you no-where.
rillian's frog
It would have to be induction with the frog being perpendicular to the field and, my intuition says, there should be some resitance to rotation. But don't flame me if I'm wrong - It's been five years since I did any physics.