Aptitude - Calendar - Discussion

Why start count odd number fron 2008 & why not start from 2007?

Please explain me.

To find the total number of odd days.

Divide the total given days by 7 and it's remainder shows odd days.

Then, Maximum number of odd days is 6.

For every ordinary year number of odd days = 1 odd day

For leap year number of odd days = 2 odd days.

If the odd days sum=7 implies odd days=0 then again loop starts from 1,2,3,4,..7 .
2008,2012,2016 are clearly leap years,

therefore
2007=1 odd day,
2008=2 (SUM=3),
2009=1 (SUM=4),
2010=1 (SUM=5),
2011=1 (SUM=6),
2012=2 odd days (SUM=8)
[HERE OBSERVE THAT SUM OF ODD DAYS EXCEEDING THE VALUE 7 THEREFORE CONTINUE ADDING THE ODD DAYS TILL WE GET A NUMBER WHICH IS DIVISIBLE BY 7],

2013=1 (SUM=9),
2014=1 (SUM=10),
2015=1 (SUM=11),
2016 = 2 odd days (SUM=13),

2017=1 (SUM=14)
upto 2017 SUM OF ODD DAYS = 14 divisible by 7 implies next year i.e.,2018 will have same calender as 2007 .

@anand ques is to find a year whose calender wil b same as 2007
so you should go for coming years where the sum of odd days will be multiples of 7(which makes exact week)
for a regular year we wil be having 365/7 which gives remainder 1 will be having 1 odd day
and for leap years we will be havin 2 odd days
YEAR ODD DAYS
2007 1
2008(LEAP) 2
2009 1
2010 1
2011 1
2012(LEAP) 2
2013 1
2014 1
2015 1
2016(LEAP) 2
2017 1
here we will get the sum of odd days as 14 which again makes exact 2 weeks...so next year will get the same calender of 2007

@All.

So, doubt here is how 14 = 0,
See,
No of days in a week is 7.
In 14 we can make to weeks,
14/7 completely divides no remainder.
Therefore 0 odd days.
All multiple of 7 will gives 0 odd days..

As per my calculations (all pl refer calendar) I have separate calculations for leap years and non leap years.

Eg: Take a leap year 2012. 5 years before this leap year and 6 years after this leap year you get the same calendar (except January and February as 2012 was a leap year and 2007 and 2018 were non leap years).

Hence you can arrive 2007 and 2018 as the same. Pl apply this logic to all leap years. I know this because I can say any days from years 1 A.D till infinity within seconds.

The year next to 1973 having the same calendar as that of 1973 is ____.

1) 1977
2) 1976
3) 1979
4) 1978

Please solve this.

How answer is 2018? Please, anyone explain me.

2009 was repeated in 2015.

2009 - 1 odd day
2010 - 1 (2)
2011 - 1 (3)
2012 - 2 (5)
2013 - 1 (6)
2014 - 1 (7)
So 2015 is the answer.

Again 2009 being non-leap year if we divide by 4 we get a remainder 1. So 2009 + 11= 2020 which is not possible 2020 being a leap year. So 2009 +6 =2015 is the repeated the year of 2009.

Why 2014 cannot be the answer. There are 7 odd days. Which sum up again to 0 odd days?