Friday puzzle -- Alice's friends

Re: Friday puzzle -- Alice's friends

Good puzzle. It'll take an extra day to figure out the answer.

Take all the time you need ;-)

Thanks, but I already did ;-)

Re: Friday puzzle -- Alice's friends

I will have to leap into action to get this one.

Too easy?

I hate to mess with a puzzle from a master like Smullyan, but I wonder if this might have been better (more difficult) if the years of birth had been 1863 or 1866 and 1867 or 1868, and the watches were next right on the March Hare's 37th birthday.

Re: Too easy?

Not necessarily. There is still some stuff you need to know. This is like many puzzles. There is something that makes it easy (or possible), but comes from something you might know. If you know or see it, the puzzle is easy, if not, then it is hard.

Re: Too easy?

Right, but the alternative removes what I suspect to be a subconscious clue for most people and requires one more fact that many people might not know.

Re: Too easy?

True, but I barely figured this one out;-)

Re: Too easy?

I think it is a pretty neat puzzle. I figured it out the hard way (20 sec difference each minute). But once I had the answer I could see there are clues in the puzzle so that the answer could be found by deduction. I think Martin Gardner called these aha puzzles.

Solution -- Alice's friends

Good job, Bill, Dan, and Larry.

The Mad Hatter is older.

Their difference between their watches increases by 20 seconds per hour, which equates to one minute per three hours or 8 minutes a day. The watches will read the same when the difference between them has increased to 12 hours. 12 hours = 720 minutes, and at 8 minutes a day, that will take 90 days. But we know it starts in January and ends in March (the March Hare's b'day). The only way that can happen is if the watches are set on noon of New Years Day, the March Hare's Birthday is 3/31, and this is a leap year. Only one of the three birth years, 1867, was 29 years before a leap year.

Larry alluded to a solution without math. The only difference between these years has to be whether or not it was a leap year. Without doing the calculation, you could figure that the watches may take too short a time to come together to be able to work on a leap year, or too long to be able to work on a non-leap year. 29 years after two of the birth years would be odd numbered years, so there are two possibilities for non-leap years, so that can't be the solution. 1967 is the only year that is 29 years before a leap year.

I originally thought this still required the calculation, since how do we know it isn't over several years that either do or do not include a leap year. Without rigorously laying out all the cases, I have convinced myself that this "meta solution" still works.

Also note that this solution begins with "The only difference between this years has to be...". That should more accurately read "The only difference I can think of between these years is...", not a basis I totally trust for a conclusion. So I would still check the arithmetic!

Re the puzzle being "easy", I think Dan hit the nail on the head. It looks easy to me now, since I see it. But to be honest, I don't remember whether it was difficult when I first saw it. I think the "29" of 29th birthday firmly planted the seed of February 29 in my mind, leading me to the solution. Thus in my harder version, I would make it the 37th birthday, and by adding 1863 as a possible birthday, I would possibly trip up some people who thought both 1863 and 1867 were 37 years before a leap year.