Friday Puzzler -- Mad Hatter and March Hare

Re: Friday Puzzler -- Mad Hatter and March Hare

david weaver

Do these watches have 12 hour faces? If so, presumably it counts if one is showing noon and the other midnight?

Re: Friday Puzzler -- Mad Hatter and March Hare

david weaver

I'll speculate at the outset before I create a possible solution that the march hare must be younger.

Here's why: This looks like a time limit type problem, and if he was older and the solution was plausible, it would also occur if he was younger. We'd have two possible solutions.

I'm assuming that the watch face must be the identical time on a 24 hour basis, not just similar looking. That creates 180 days between times that the watches show the identical time. At that point, one watch has gained a day and one has lost a day. But we don't care what date the watches would show, just the time.

Re: Friday Puzzler -- Mad Hatter and March Hare

david weaver

The maximum time between January 1 on a non-leap year - at noon, and march 31 at 11:59 is 89.5 days.

I made a little mistake on the other page. It won't be noon when the watches match time again, so it won't be exactly 180 days - closer to 179.

Re: Friday Puzzler -- Mad Hatter and March Hare

Larry Barrett

Each watch gains (loses) 240 seconds = 4 minutes each day, or 2 hours every 30 days.

From noon on Jan 1 to noon (actual time) on March 31 is 89 days in a non-leap year, or 90 days in a leap year. So in a leap year, on March 31, one watch will read noon + 6 hours = 6 PM and one watch will read noon - 6 hours = 6 AM. Although the instructions are not clear, one interpretation is that at this instant the watches read the same.

If the March Hare was born on March 31 1867, then it will be 29 years old on March 31, 1896, which is a leap year.

So I think 1867 is when the March Hare was born and the Mad Hatter was born in 1866.

Re: Friday Puzzler -- Mad Hatter and March Hare

Alex Y

the two of them were showing each other their watches (which are normal, not like the one alluded to in Carroll).

I would contend that a "normal" watch in the 19th century would have 12-hr dial.

Correct!

Alex Y

After 90 days, both watches will be identical--big hands on 12 and little hands on 6. That's the time the "normal" watches show; am or pm is in the minds of the watch owners.

Re: Correct!

david weaver

bummer, I was working though 180 days yesterday and getting too long of a time span.

And then it was solved before i got back, but as I suspected since it's a time limit problem, the hare had to be younger or there would be more than one solution which would be in bad taste!!

Time limit problem?

Alex Y

David, you completely lost me there. I would be interested if there is another approach. And what does it mean that there are two solutions if the March Hare is older? Please elaborate.

My approach was the same simple arithmetic that Larry did. (The arithmetic is simple--not so for finding the leap year role, although the puzzle has an embedded hint with the 29th birthday.) The only shortcut I saw after the fact is that knowing that the March Hare's 29th birthday is in a leap year, we know that his birth year was an odd number. I'd still want to check myself by confirming that 29 years after 1867 was a leap year.

Re: Time limit problem?

david weaver

I forgot to respond to this. My apologies for being slow - I think I made a mistake in my thinking about the time limit problem issue as I wasn't considering leap years, just assuming they didn't apply.

Re: Time limit problem?

Alex Y

Aha! Yes, the leap year is the key. And the puzzle has a subtle hint that might lead you there, in that this is his 29th birthday, while the puzzle would have exactly the same result if it were his 21st, 25th, 33rd, ... birthday.