In 1952, Professor Egghead's father voted for Dwight Eisenhower for US President and had kept the calendar as a keepsake. Recently, Professor Egghead was going through some of his father's keepsakes and discoverd this calendar from 1952 and noticed that all the dates in 1952, which was a leap year, coincided exactly with the dates in 2008, which is also a leap year.
This made the professor wonder, how many different calendars would you need to have in order to represent every possible combination of yearly calendar?
Monday, February 11, 2008
Subscribe to:
Post Comments (Atom)
3 comments:
In a normal year there would be 52 weeks and 1 extra day, so in leap years there would be two extra days.
Year = which day it starts on
_____________________________
Year 1= monday
Year 2= tuesday
year 3= wednesday
Year 4= thursday (leap year)
Year 5= saturday
Year 6= sunday
Year 7= monday
Year 8= tuesday (leapyear)
Year 9= thursday
Year 10= friday
Year 11= saturday
Year 12= sunday (leapyear)
Year 13= tuesday
Year 14= wednesday
Year 15= thusrday
Year 16= friday(leap year)
Year 17= sunday
Year 18= monday
Year 19= tuesday
Year 20= wednesday (leap year)
Year 21= friday
Year 22= saturday
Year 23= sunday
Year 24= monday (leap year)
Year 25= wednesday
Year 26= thursday
Year 27= friday
Year 28= saturday (leap year)
Year 29= monday
This proves that you only need 28 different calenders.
The reason I have put monday on a 29th year is because that is when the wole cycle starts again.
By the way, I am only 13.
well since you skip leap years there may be some problems with this theory bug sure.
You need 14 calendars. The puzzle didn't say anything about the calendars needing to represent consecutive years. You need 7 calendars from a leap year, and 7 from a non-leap year, with the LY calendars having Jan 1st on M/T/W/H/F/S/U (and the same for the non-LY calendars).
Post a Comment