20070708, 01:21  #1 
"Jason Goatcher"
Mar 2005
110110110011_{2} Posts 
Day Million
I don't know if this would count as a puzzle, but...
I read a science fiction short story a few years ago that tried to predict what life would be like on the millionth day of our calendar(I believe it's called the Gregorian calendar?) Including all the leap year rules, and remembering that there was no 0 year, what is the millionth day of this calendar. Has it already happened? I can't seem to get it to add up, I'm not sure why. 
20070708, 02:10  #2 
Aug 2002
Buenos Aires, Argentina
2×3^{2}×83 Posts 
Dividing 1000000 by 365.25 in order to find an approximate year we obtain 2737.85.., which is near 2738, so the date should be near 1/1/2739.
So now we will find the number of days between 1/1/1 and 1/1/2739 and then adjust. First we will need to know how many leap years are between these two dates. There are 2738/4 = 684 years multiple of 4, but 27 of them are multiple of 100 and 6 multiple of 400. So we have 68427+6 = 663 leap years. So we have 365*2738+663 = 1000033 days between 1/1/1 and 1/1/2739. Between 1/1/1 and 12/1/2738 we have 1000033  31 = 1000002 days. Between 1/1/1 and 11/29/2738 we have 1000002  2 = 1000000 days. So the millionth day of the Gregorian Calendar is November 28th, 2738 
20070708, 02:13  #3 
Jun 2003
3·5·107 Posts 
What about Feb 30 th? There are only 97/400 leap years http://en.wikipedia.org/wiki/Gregorian_calendar
Also isn't the first day 1/1/00.? The day today is 732864. (based on wiki URL) You can play with this http://www.abdicate.net/cal.aspx to get the gregorian serial date Last fiddled with by Citrix on 20070708 at 02:32 
20070708, 02:16  #4 
Aug 2002
Buenos Aires, Argentina
2726_{8} Posts 
I've never heard about February 30th. The first year is the number 1. There is no year zero.
The date you showed above does not count the nonexistent year zero. Using the same idea we will compute first the number of leap years: the number of multiples of 4 is 2006/4 = 501, but there are 20 multiples of 100 and 5 multiples of 400, so there are 50120+5 = 486 leap years. So the number of days elapsed between 1/1/1 and 1/1/2007 is 365*2006+486 = 732676. Between 1/1/1 and 2/1/2007: 732676+31 = 732707 Between 1/1/1 and 3/1/2007: 732707+28 = 732735 Between 1/1/1 and 4/1/2007: 732735+31 = 732766 Between 1/1/1 and 5/1/2007: 732766+30 = 732796 Between 1/1/1 and 6/1/2007: 732796+31 = 732827 Between 1/1/1 and 7/1/2007: 732827+30 = 732857 Between 1/1/1 and 7/7/2007: 732857+6 = 732863 So today, July 7th 2007 is the 732864th day of the Gregorian calendar as you state above. Last fiddled with by alpertron on 20070708 at 02:51 
20070708, 03:45  #5  
Cranksta Rap Ayatollah
Jul 2003
1010000001_{2} Posts 
Quote:


20070708, 03:54  #6  
Oct 2006
100000100_{2} Posts 
I found the website http://www.timeanddate.com/date/dateadd.html which gives what alpertron worked out for himself (I'm just too lazy ):
Quote:


20070708, 07:48  #7  
Bronze Medalist
Jan 2004
Mumbai,India
2052_{10} Posts 
Quote:
Just the type of maths to read on a sunday morning especially for an arm chair math'cian like myself Mally 

20070708, 13:57  #8 
Aug 2002
Buenos Aires, Argentina
2·3^{2}·83 Posts 
The Gregorian Calendar was implemented in some countries first in 1582, but it does not mean that there are no dates prior to October 15th, 1582 in this calendar. You just use its rules. One of them is that there is no year zero.

20070709, 00:52  #9  
Cranksta Rap Ayatollah
Jul 2003
641 Posts 
Quote:


20070817, 23:54  #10 
Oct 2005
44_{8} Posts 
Do any of these calculations take into account the 'lost' ten days in October 1582 when the Gregorian calendar was introduced?
Do they need to? MS63 
20070820, 20:01  #11  
Jul 2006
USA (UT5) via UK (UT)
354_{8} Posts 
Quote:
The JD corresponding to 1 Jan. 1 is 1721423.5. If you count this as Day 1, then the 1 millionth day after this is JD 2721422.5, which corresponds to 2738 Nov. 26. If you count 1 Jan. 1 as Day 0, then the 1 millionth day is JD 2721423.5, which corresponds to 2738 Nov. 27. Gareth Williams 

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