Aptitude - Calendar - Discussion

My best answer is hear:

1. Odd days:

2007 have 365 days (2007/4 = 3 (not perfect divisible) so it is not LEAP year and having 365 days. Now divide by 7 (number of weeks) to know how many weeks in 2007.

365/7 = 52.1 so 52 weeks and 1 day. So the year 2007 have 52 weeks and 1 day. This 1 day is consider as odd day).

2. So 2007 having 1 odd day.

2008/4 = 0 so LEAP year contain 366 day i.e. 52 week 2 odd days.

Similarly 2009 2010 2011 2012 2013 2014 2015 2016 2017.

1 2 1 1 1 2 1 1 1 2 1.

3. Now add all odd years because the year is repeated when odd days = 0.

4. 1+2+1+2+1 = 14.

5. 14 means it is perfect 2 weeks so no odd days.

6. We not get 7 in adding so we consider 14 (perfect week is needed).

7. 2018 is not included because upto 2017 it is 14 so next year is repeated year.

As per my calculation (Please refer calendar if you want to verify). For leap years there is one calculation. So having leap year as a base 5 years before the leap year and 6 years after the leap year you get the same calendar (Except January and February) as the base year considered here is a leap year.

Eg. 2012 was a leap year. 5 years before 2012 i.e. 2007 had the same calendar as 2012 and 6 years after 2012 that s 2018 are the same. Since 2012 was a leap year Jan and Feb would have differed from 2018 and 2007 from 2007 as we all know leap year effects are observed only in Feb where one day is added in Feb.

So hereby we conclude that 2007 is the same as 2018 as given in the question. I case any days from years 1 A.D till infinity.

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

It is so simple to calculate , no need to find Jan 1, 2007.

If the year is completely divisible with 4,

i.e remainder = 0 then simply add 23 years to the question year.
remainder = 1 then simply add 6 years to the question date,
remainder = 2 or 3 then simply add 11 years to the question.

Example:

Which year is same as;

2004 ? answer : 2004 + 23 = 2027.
2005 ? 2005 + 6 = 2011.
2006 ? 2006 + 11= 2017.
2007 ? 2007 + 11 =2018.

2004, 2008, 2012............ +23.
2005, 2009, 2013..............+ 6.
2006, 2010, 2014..............+11.
2007,2011, 2015...............+11.

Is there any mistakes excuse me.

Whenever they ask for the same calender we should check whether it is a leap year or not. If it is a leap year add 28 to the given year then you will get the same calendar for leap year for non- leap years. 6 and 11 (6 is added when given year is subtracted from previous leap year u shoud get 1.

i.e ex=2005 is not a leap year..previous leap year is 2004 so (04-05=1), then add 6 because we got 1 after subtracting non-leap year with previous leap year so (2005+6=2011).

(11 is added when you subtract non-leap year to previous leap year you will get 2 nor 3).
Ex:2006 is a non-leap year, previous leap yr is 2004 so(04-06=2).
i.e 2006+11=2017.

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.

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.

If the year is leap then add 28 to get the same calendar provided that no century year which does not leap should fall in between.

If the year is not leap, add 11 and if the result is not leap; the result has the same calender. If the result is leap then add 6 to get same calender. It should be taken that no century year which is not leap should not fall in between.

2007 is not leap so 2007+11=2018, 2018 is not leap; hence 2018 is the answer.

@All

Given;

Year +6
2007+6=2013.

Find out how many leap year there in between 2007 and 2013.

there is 2008 and 2012 (2 Leap year).

If there is 2 Leap year then add 11 to the given question. It will be your answer.

2007+11 = 2018.
Instead of;
If there will be 1leap year while adding 6 in given question the answer will be given year +6= answer.

If in question there will be Leap year. Then add +5 it will be your answer.