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Twins birthday puzzle


bonanova
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A difference of two days (1 in between) is easy to explain ...

At the the time the mother of the twins went into labor, she was traveling by boat, plane or ambulance and crossed (any) time zone in the process . The older twin was was born just before midnight on March 1st of a non-leap year, followed by the second just after midnight (5 minutes later) on 28 February after crossing the time zone. During a leap year the younger twin celebrates her birthday two days before her older twin.

But, could there ever be a time-zone difference of more than 24 hours to make one twin celebrate three days after the first (i.e. two days in between according to the OP)? Unless traveling was in space during labor. Otherwise, the twins decided not to celebrate their birthdays on the same day!

 

Spoiler

I am changing " (any) time zone" to the International Date Line since "(any) time zone" won't work!

 

Spoiler

Crossing will be from West to East

 

Edited by rocdocmac
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Rocdocmac is close. You get one day for the 5 minutes, one day for crossing the international date line, and one day for a leap year.

So if you are east of the international date line at 11:59 on the 28th of February, then cross the line and wait a minute or so, it becomes the 2nd of March. On a subsequent leap year these  dates could correspond to Sunday and Wednesday.

 

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6 hours ago, kaspar said:

Rocdocmac is close. You get one day for the 5 minutes, one day for crossing the international date line, and one day for a leap year.

So if you are east of the international date line at 11:59 on the 28th of February, then cross the line and wait a minute or so, it becomes the 2nd of March. On a subsequent leap year these  dates could correspond to Sunday and Wednesday.

Nice solve. And the "few years back" is actually two years ago, when Feb 28 was Sunday and March 2 was Wednesday.

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