YABS # 3 - to torture you further

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borogovio

1
This one is not as bad as the one about the three daughters I posted recently:

It's about the Hare of March and the Mad Hatter. One of the was born in 1842 - but I won't tell you which -, and the other one in 1843 or in 1844 - but again, I won't tell you which -. But obviously the Hare of March was born in March.

The Hare of March has a watch that goes 10 seconds ahead each hour. The Mad Hatter has a watch that goes 10 seconds back each hour. (*) One day of January they both set their watches to the right time at exactly noon.
<< Are you aware >>, said the Mad Hatter, << that our watches won't show the same time until your next birthday, when you become 21 years old? >>
<< Perfectly aware >>, replied the Hare of March.

Question: Who is older: the Hare of March or the Mad Hatter?
(As usual, the answer has to be justified !)


(*) As I'm not sure if I have translated it right, I'll explain a little further. If both watches are set to 12 just at noon, 1 hour later the watch of the Hare of March will show 01:00:10 whilst the watch of the Mad Hatter will show 00:59:50; 2 hours later, 02:00:20 and 01:59:40, respectively, and so on.
 
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borogovio

1
Some more days have passed - no more solutions, so here comes the good one, slightly modified from teutonu:

The Mad Hatter is older, as the hare was born on the 31 march 1843. Here's the argumentation:
If the watches are running "away" 20 secs per hour this means that they need 2160 hours to be again even. This is exactly 90 days. We are looking for the birthday of Hare which is in march, that means that they set their watches on the 1 of January, else it won't fit until the end of march. And also it won't fit if you only have 28 days in february, so it must be a leap year to have his "next birthday" in march. 1864 was a leap year. And only 1843 + 21 gives 1864, so Hare was born in 1843. Mad Hatter was born 1842 and is therefore older.

This problem comes from 'Alice in Puzzle-land', by Raymond Smullyan
 
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