For many years,
the Roman Catholic church used the
Julian calendar, which has a leap year
every year that is divisible by four,
making the average calendar year
longer than the sidereal year and
causing the date of the first day of
spring to change gradually.
To correct this, Pope Gregory decreed that
Thursday, October 4, 1582, would be
followed by Friday, October 15. He
also declared that years divisible
by 100 would be leap years only if
divisible by 400.
For any year since
1582, if one printed two 12-month
calendars, one Julian and the other
Gregorian, with dates for the days
of the month, at least some of the
dates would not fall on the same day
of the week.
What is the first year
for which each day of each month will
fall on the same day of the week for
both calendars?
In 1592, the day that would have been October 5, per the Julian calendar, became October 15, a 10 day difference. We need a 14-day difference. The differences caused by the skipping of a leap year are in the same direction as the 10-day skipping, so four more skipped leap years need to occur: 1700, 1800, 1900 and 2100.
The skipped leap-year day in 2100 would be in February, so January and the early part of February would not match in the two calendars. Therefore 2101 will the the first year in which the two calendars match the entire year, as far as date vs day of week.
The verify: Checking a program that does conversions, December 22, 2101 Gregorian is December 8, 2101 Julian, and 2101 is not a leap year in either, so the offset is 14 days, two full weeks.
January 30, 2100 Gregorian is January 17, 2100 Julian, so it's not an integral multiple of 7 days off, as it's before the left-out leap year day.
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Posted by Charlie
on 2025-01-14 08:55:25 |
solution
