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Old 2007-07-08, 01:21   #1
jasong
 
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Default 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.
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Old 2007-07-08, 02:10   #2
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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 684-27+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
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Old 2007-07-08, 02:13   #3
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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 2007-07-08 at 02:32
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Old 2007-07-08, 02:16   #4
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Quote:
Originally Posted by Citrix View Post
What about Feb 30 th? Should it be Nov 26th, 2738
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 non-existent 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 501-20+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 2007-07-08 at 02:51
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Old 2007-07-08, 03:45   #5
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Quote:
Originally Posted by alpertron View Post
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 non-existent 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 501-20+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.
There's also no year 1581 in the Gregorian calendar, since it was implemented in 1582, so what's the meaning of a "year zero"?
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Old 2007-07-08, 03:54   #6
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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:
From date: Saturday, January 1, 0001 (Julian calendar)
Added 1,000,000 days

Resulting date: Sunday, November 27, 2738 (Gregorian calendar)
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Old 2007-07-08, 07:48   #7
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Quote:
Originally Posted by alpertron View Post
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 684-27+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
Brilliant alpertron, brilliant!.

Just the type of maths to read on a sunday morning especially for an arm chair math'cian like myself

Mally
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Old 2007-07-08, 13:57   #8
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Quote:
Originally Posted by Orgasmic Troll View Post
There's also no year 1581 in the Gregorian calendar, since it was implemented in 1582, so what's the meaning of a "year zero"?
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.
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Old 2007-07-09, 00:52   #9
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Quote:
Originally Posted by alpertron View Post
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.
http://en.wikipedia.org/wiki/Astrono...year_numbering
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Old 2007-08-17, 23:54   #10
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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
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Old 2007-08-20, 20:01   #11
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Quote:
Originally Posted by alpertron View Post
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 684-27+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
Why not use the Julian Day numbers to do this calculation?

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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