k-man

I found a flaw in my reasoning, so I will have to think about it some more later...

Your result is 1116.833333. I tried similar approaches and could not even get to 2000. So, kudos to Nana7 for the quick answer and to superprismatic for verifying it. I would like to know how Nana managed to get this answer?

Nice puzzle, xamdam. Kudos to Drexlin, who posted this answer first in post #8.

EDIT: I see that Phil beat me to it...

The OP states that their years are equivalent.

Not sure what you mean by generalizing it, but a proof is pretty straight forward.

It's impossible to reach 12345...BACDE from 12345...ABCDE by following the rules as others have already mentioned. You must have made a mistake somewhere along the way. Getting from 12345...BACDE to ABCDE...12345 is pretty straight forward. Just follow the steps in kevink's post.

My initial thinking was the same as Nana's and I don't see how this solution violates any of the OPs constraints. No matches need to be burned and the resulting shape has the desired area. Oh, and while I like Mukul's solution this one requires no calculations and can be performed by a child.

My guess...

Your proof assumes a particular method of sending letters. You have proven that using this method of sending letters it takes 2(n-1) steps for any n. That is not what required proof, though. We need to prove that there is no such method that would complete the task in fewer than 2(n-1) steps. Therefore, a proof cannot assume any particular method of sending letters.

@Cube This has been posted a number of times before. Please search the forum before posting new riddles. A simple search for "ice puddle" or "water rope" will find the previous posts of this riddle.

perfect! Um... not quite... If the suggestion is to discard these 3 pills then there isn't enough pills to survive 29 days. Not to mention that crossing the international date line merely changes the date, but won't add a day to your life.

LOL

I agree with you that the puzzle could've been worded better and some ambiguity could be resolved, but nonetheless, there were some clarifications made later that made the puzzle solvable. It was clarified that husband and wife do not shake hands. So, among 8 people made up from 4 couples, the maximum number of handshakes one person can perform is 6 and the minimum is 0. The logic of coupling (6,0), (5,1), (4,2) and (3,3) is this: Let's call the person who shook 6 hands Amy. Amy shook hands with everybody except her spouse, so everybody except Amy's spouse shook at least one hand - Amy's hand. This leaves only Amy's spouse as the person who shook 0 hands. Let's call the person who shook 5 hands Bill. Bill shook hands with everybody except Amy's spouse and his own spouse. Now, everybody except Amy's and Bill spouses shook at least 2 hands - Amy's and Bill's. This leaves Bill's spouse as the only person who shook 1 hand - Amy's. The same can be continued further to couple 4 with 2 and 3 with 3. So if one knows how many hands he/she had shaken, one can deduce how many hands one's spouse had shaken. P.S. Given that the both the OP and amitgol didn't use spoilers I didn't use one either.