Which of the following years will have the same calendar as that of 2007?
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An interesting followup question is "Which calendar will be the same as 2016?" Can you post that version too?
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As 2016 is a leap year so we can add year = LCM of 7 and 4 ( 7*4=28) {nmbr of day in a week and leap year frequency} so 2016 + 28 = 2044
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I don't think "adding LCM of 7 and 4" is the right way to express the idea.
Or, at least, more explanation is required like:
1. It has to be a multiple of 4
2. Every 4 years (with the exception of multiples of 100/400/etc), we add 5 days
3. Number of days added has to be a multiple of 7
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@Calvin Lin – OK, but you asked for repeating calendar of a leap year that only repeats in every 28 years. If you have better logic for this "28" you can share with me.
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@Vipendra Singh – For example, since 1900 is not a leap year, hence if we worked with the year 1896, then the next identical calendar will not be 1896+28.
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@Calvin Lin – Yaa you are right but while entering 1896 to 1896+28= 1924 century is changing ( from 18th to 19th Century) so in this case we have to add 40 years. SO next same year as 1896+40= 1936
As 2016 is a leap year so we can add year = LCM of 7 and 4 ( 7*4=28) {nmbr of day in a week and leap year frequency} so 2016 + 28 = 2044
You should have mentioned 2012 to confuse the people as the method I used first lands up on 2012 but is ignored as it is a leap year
Let Jan 1, 2007 Monday be day 1 or D ( 2 0 0 7 ) = 1 and y years or d days later the calendar is the same. Then Jan 1, 2007 + y is D ( 2 0 0 7 + y ) = 1 + d . If year 2 0 0 7 + y has the same calendar, then Jan 1, 2007 + y has the same weekday Monday. Using modular arithmetic, we say that since D ( 2 0 0 7 ) = 1 ≡ 1 (mod 7) ⟹ D ( 2 0 0 7 + y ) = 1 + d ≡ 1 (mod 7) ⟹ d ≡ 0 (mod 7) .
We note that for normal year 3 6 5 ≡ 1 (mod 7) and leap year 3 6 6 ≡ 2 (mod 7) . And that 2008 is a leap year, we conclude that for y ≥ 2 :
d ≡ ( y + ⌊ 4 y − 2 ⌋ + 1 ) (mod 7) The leap day is added at the end of the year, hence − 2
⟹ y + ⌊ 4 y − 2 ⌋ + 1 y + ⌊ 4 y − 2 ⌋ + 1 ≡ 0 (mod 7) = 7 n where n is a positive integer.
The year 2007 has 365 days and starts on a Monday. So, we need to identify what will be the next not leap-year that will start on a Monday. We know that every year we start one day after (Ex: 2008 starts on a Tuesday). On a leap-year, we advance two days. So, the next year that will starts on a Monday is: - 2008: Tuesday - 2009: Thursday - 2010: Friday - 2011: Saturday - 2012: Sunday - 2013: Tuesday - 2014: Wednesday - . - . - 2017: Sunday - 2018: Monday
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A normal year has 52 weeks and one additional day. This additional day is called an 'odd day' as it is the left over and will be carried forward to the subsequent week of the next year. A leap year has 2 odd days. Let 2007 starts with Monday. Then 2008 will start from Tuesday (add 1 odd day). Like this 2009 will start from Thursday (add 2 odd day as 2008 was a leap year) an so on ( 2010 - Fri, 2011 - Sat , 2012 - Sun, 2013 - Tue { 2 odd in 2012} , 2014- Wed, 2015- Thu, 2016 - Fri , 2017- Sun { again as 2016 is a leap year so add 2 odd day } , 2018 - Mon. Thus 2018 also starts from Monday and also a Gregorian calendar ( non leap year 365 day ) so calendar of 2007 and 2018 will be same.