# Aptitude - Calendar - Discussion

### Discussion :: Calendar - General Questions (Q.No.11)

11.

The calendar for the year 2007 will be the same for the year:

 [A]. 2014 [B]. 2016 [C]. 2017 [D]. 2018

Explanation:

Count the number of odd days from the year 2007 onwards to get the sum equal to 0 odd day.

```Year    : 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017
Odd day : 1    2    1    1    1    2    1    1    1    2    1
```

Sum = 14 odd days 0 odd days. Calendar for the year 2018 will be the same as for the year 2007.

 Bhuvana said: (Jun 4, 2010) Please explain me that how 14 odd days is similar to 0 odd days

 Mani said: (Jun 27, 2010) How to find odd days? and how it's possible?

 Mani said: (Jun 27, 2010) Need a brief explanation for this problem?

 Mani` said: (Jun 27, 2010) I don't have much time to wait for your explanation, It's very emergency ? so, please respond to my questions ?

 Harru said: (Jul 2, 2010) Sum = 14 odd days 0 odd days. plz clerify this point

 Sundar said: (Aug 18, 2010) Hi Friends, Please go through the basics terms and formulas given below before solving the problems. Link: http://www.indiabix.com/aptitude/calendar/formulas Hope this will help you to solve the problems in this section.

 Nitesh Nandwana said: (Oct 10, 2010) The number of days more than the complete weeks are called odd days.

 Prasad said: (Dec 23, 2010) 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 .

 Manisha said: (Jan 6, 2011) You are right prasad

 Priya said: (Mar 22, 2011) This calendar method will take time to find the day of the week.... I have seen a new easy method in an quantitative aptitude book BEACON....if possible read that book

 Hariharan said: (Jun 15, 2011) Hi All Pls explain me why 2018 is not included in this step.

 Jati N Kamani said: (Jul 7, 2011) Please explain me how to 14 odd days to equal 0 odd day.

 Shahir said: (Jul 14, 2011) We are supposed to find the day of the week on a given date. For this, we use the concept of 'odd days'. In a given period, the number of days more than the complete weeks are called odd days.

 Akash said: (Jul 27, 2011) Can anyone explain how we are counting the days in respect to the odd days?

 Madhu said: (Jul 30, 2011) @Akash Actually 7 odd days = 0 odd days right?? And, even for 2012 2 odd days are assigned not for 2011 as the former is divisible by 4...

 Prem said: (Aug 3, 2011) Just simple every 12 month once calender remain same. Am I correct ?

 Rehan said: (Aug 5, 2011) Prasad is absolutely right.

 Asmath said: (Aug 22, 2011) Prasad you are absolutely correct.

 Riya said: (Aug 24, 2011) Hii prasad. I have an confusion, Let suppose I have an year x and want to know the next same year, if I get sum of odd days 7 in this case => odd days=0, Now what is the answer? 1) is the next year will be the same as x as you told in your comment?

 Anand said: (Aug 26, 2011) I have query regarding year 2002. Years- 2002, 2003, 2004, 2005, 2006, 2007, 2008, 2009, 2010, 2011, 2012, 2013. No of odd days for respective years. 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2. Till 2007 total odd days becomes 0 hence answer should be 2008 but it is 2013 why ?

 Sandy said: (Aug 29, 2011) The year 2002 calender and the year 2007 have the same calender.

 Apbakshi said: (Sep 2, 2011) @ sandy. Will you tell me the calculation behind that ?

 Svetta said: (Jan 22, 2012) Wow prasad! nice.

 Rambabu said: (Mar 1, 2012) Please briefly explain this problem.

 Mahesh said: (Mar 22, 2012) Please tell us how to findout a day when date, month, year are given.

 Ramu said: (Aug 28, 2012) @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

 Shashi said: (Oct 2, 2012) The calender for the year 2008 will be the same for the year.

 Upendra said: (Sep 15, 2013) Guys calender will be same if their number of odd days will be same 2007..6*1+2=8,1 odd day. 2014-3 odd days. 2017-0 odd days. 2018-1 days. So answer will be 2018.

 Poornimajagatha said: (Sep 20, 2013) As in formulas we heard that for 1990 = 1600+300 => 0+1 => 1(odd day) => is that sunday or monday?

 Prabhuchandu said: (Jan 22, 2014) Easy method: 2000+28 = 2028. 2001+6 = 2007. 2002+11 = 2013. 2003+11 = 2015. 2004+28 = 2032. 2005+6 = 2011. 2006+11 = 2017. 2007+11 = 2018.

 Prabhuchandu said: (Jan 22, 2014) Easy method: 2000+28 = 2028. 2001+6 = 2007. 2002+11 = 2013. 2003+11 = 2015. 2004+28 = 2032. 2005+6 = 2011. 2006+11 = 2017. 2007+11 = 2018.

 Sundar Gs said: (Jan 31, 2014) @Prabhuchandu. Can you explain why you add years with 6, 11, 28...and all?

 Jenny said: (Jun 19, 2014) Why we want to find the odd numbers only from 2007 to 2017. Why we not find the odd numbers 2018?

 Chandu said: (Sep 4, 2014) How can you tell that we should add 6, 11, 28? Can you explain it.

 Amol Jadhav said: (Sep 26, 2014) @Prabhuchandu. Can you explain why you add years with 28, 6, 11, 11..and all?

 Shreyas D.J. said: (Oct 14, 2014) The addition is coming 14 odd days = 0 odd days for year 2017 not 2018. So why take 2018 if it has 1 odd day?

 Gopi said: (Dec 24, 2014) Please explain about 2018 why we are not considering 2018? Please explain quickly very urgent.

 Rya said: (Jan 21, 2015) Please explain clearly how that 14 odd days is equivalent to 0 days?

 Katy said: (Feb 28, 2015) X/4 (x=given year). For any year divided by 4, the possibility of remainders are 0, 1, 2, 3. If remainder is 0---->x+28. If remainder is 1---->x+6. If remainder is 2 (or) 3------>x+11. So 2007 divide by 4 gives 3 as remainder. So 2007+11 = 2018.

 Rupali said: (Mar 9, 2015) Why we are not considered 2018? Answer please.

 Akriti said: (Apr 2, 2015) Why we are not considering 2018? According to me it should be 2017 as both 2007 and 2017 has 0 odd days. Please explain.

 Murali.O said: (Jul 6, 2015) Hai friends one small way is to add the number in 11 ok.

 Master Black said: (Oct 28, 2015) Can you just calculate the next same year for 2008?

 Paveek said: (Nov 15, 2015) 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.

 Paveek said: (Nov 15, 2015) @Hariharan. 2018 is not included because upto 2017 it is 14 so the next year is repeated year. We just need perfect week (7, 14, 21....) next year of the perfect week is the repeated year.

 Awdhesh said: (Dec 1, 2015) Explanation is not considered clearly. Find easy method please?

 Sudarshan said: (Dec 16, 2015) 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.

 Sudarshan said: (Dec 16, 2015) 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.

 Sudarshan said: (Dec 16, 2015) 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.

 Shekharrao said: (Dec 31, 2015) What is mean of ordinary year?

 Rakesh said: (Jun 2, 2016) How can we decide that the year is an ordinary year and leap year? Please help me.

 Saravanakumar said: (Jul 4, 2016) It is as same as 400th year (leap year) can any one differenciate it?

 Jesu said: (Jul 12, 2016) To find that the year is a leap year or an ordinary year. The year should be divided by 4 if you get the remainder as 0 then it must be a leap year or else if you get a remainder as 1 or 2 or 3 then it must be a ordinary year. Hope you understand.

 Gurubalaji said: (Jul 15, 2016) If you get a equal calendar for x year= (x + 11) year. Because equal calender's obtain in every after 11 year.

 Guru Bhaskar Reddy said: (Jul 16, 2016) 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.

 Pradeep Verma said: (Jul 17, 2016) Please tell me the solution for this problem. If, 24march 1992 is Friday then which day fall on 24september 1993?

 Rakesh said: (Aug 10, 2016) It's for the upcoming years. How to calculate the same year backward?

 Venkatesh said: (Sep 4, 2016) 2007 is a normal year and front year 2006 is also a normal year then we have to add 11years to the given year. i.e; 2007 + 11years = 2018.

 Ramandeep said: (Oct 8, 2016) Simply divide 14 by 7 if zero remainder so odd day zero.

 Pavan Kumar said: (Oct 16, 2016) Thanks for the explanation. I have a small doubt. As per your method for calendar 1988 year the calendar should be repeated on 1992. But it is repeated on 2016. How it is possible? Please explain.

 Rakesh Parmar said: (Nov 12, 2016) @All, Simply remember this. Divide Any year by 4. If you get 0, add 28 to given year, Ex. 2000+28 = 2028. 2000 and 2008 both same year. If you get 1 add + 6. If you get 2 or 3 add + 11.

 Sonali said: (Dec 9, 2016) Thanks @Prasad.

 Santhosh said: (Jan 3, 2017) I have a query regarding the year 2006. No of odd days for respective years. 1, 1, 2, 1, 1, 1. Till 2011, total odd days becomes 0 hence answer should be 2011 but it is 2017 why?

 Santhosh said: (Jan 3, 2017) I have a query regarding the year 2006. No of odd days for respective years. 1, 1, 2, 1, 1, 1. Till 2011, total odd days becomes 0 hence answer should be 2011 but it is 2017 why?

 Santhosh said: (Jan 3, 2017) I have a query regarding the year 2006. No of odd days for respective years. 1, 1, 2, 1, 1, 1. Till 2011, total odd days becomes 0 hence answer should be 2011 but it is 2017 why?

 Navnath said: (Jan 6, 2017) 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.

 Mani said: (Jan 7, 2017) Thank you very much. I understand now.

 Sumanth said: (Jan 22, 2017) It is 1979 @Navnath.

 Niks said: (Feb 9, 2017) Thank you for easy method @Prabhuchandu.

 Krishna Kittu said: (Feb 17, 2017) For finding repeated calendars we used formulae of adding 28, 6, 11. for finding the past repeated calendars we have to subtract same 28, 6, 11. Am I right?

 Riya Khandelwal said: (Apr 10, 2017) I am not understanding this. Please explain how 2018 will be similar to 2007?

 Siri said: (Apr 21, 2017) 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.

 Aishwarya said: (Jun 5, 2017) The number of extra days more than a complete week in a given period is called odd days. Hence there are 14 days which can be of 2 weeks. As 7 days is one week and other 7 days are another week so there are no more days which are beyond a complete week. Therefore, the odd days are 0.

 Sowmya said: (Jun 8, 2017) Is this procedure is same for leap year? Please, anyone explain.

 Balu Srinivas Reddy said: (Jul 13, 2017) For 4n => add 28 years. If 2004 Will same in 2004+28=2032 For 4n+1 => add 6 years. 4n+2 & 4n+3 - add 11 years.

 Vijay said: (Aug 9, 2017) Please explain in detail.

 Anu said: (Aug 15, 2017) It applies to all years: Leap year == add 28. 1 yr after leap year== add 6. 2 yr after leap year== add 11. 3 yr after leap year == add 11. Here 2007, nearest leap year before 2007, 2004. Therefore 2007 is 3 yrs after 2004. Therefore add 11 to 2007=2017.

 Manjunath N said: (Aug 27, 2017) Assume 2007 starts with Sunday. 2008 starts - monday. Since 2008 is leap year hence 2009 starts - wednesday. 2010 starts - thursday. 2011 starts- friday. 2012 starts-sat. 2012 is leap year 2013 starts- monday. 2014 starts - tuesday. 2015 starts- wednesday. 2016 starts- thursday. 2016 is leap year 2017 starts- saturday. 2018 starts - sunday Hence 2018 will have same year as 2017.

 Pratik said: (Sep 19, 2017) How to find answer this question? Seema remembers that her mother's birthday falls on after 21st January but before 25th January while his brother remembers that it falls after 23rd January but before 28th January. When is Seema mother birthday?

 Sharayu said: (Sep 24, 2017) How is the odd days for year 2017 1? I still didn't get this?

 Prudhviraj said: (Oct 17, 2017) @ALL. start with following year 2008 ->2 09 ->1 10 ->1 11 ->1 12 ->2 ------- sum= 7 is divisible by 7 (no option for 2012) -------- 13 ->1 14 ->1 15 ->1 16 ->2 17 ->1 18 ->1 ------- sum= 7 is divisible by 7 ( option D for 2018 ) -------

 Jitendra Singh said: (Nov 22, 2017) Calander of 2016 is repeated in which year? Please explain.

 Dfeverx said: (Apr 11, 2018) For every ordinary year number of odd days = 1 odd day. For leap year number of odd days = 2 odd days. 2016-2 (sum=2) 2017-1 (sum=3) 2018-1 (sum= 4) 2019-1 (sum=5) 2020-2 (sum=7) It implies next year i.e 2021 will have the same calendar as 2016.

 Shivam Giri said: (May 4, 2018) The calendar of 1872 repeated on which year and why? Please explain the answer.

 Rupa said: (Jul 1, 2018) Why this trick is not applicable to 2008. Can anyone explain?

 Vinod said: (Jul 5, 2018) Nice, Thanks @Prasand.

 Evlin said: (Aug 19, 2018) Good one @Katy. It's a time efficient method. Thanks.

 Jitendra said: (Sep 28, 2018) 2007 last two digit division by 4. if the remainder is 1, add 6. remainder is 2 add 11, and 3 add 11 get an answer.

 Akanksha said: (Sep 29, 2018) Why do we take till 2017? Please explain.

 Semran Sheikh said: (Oct 9, 2018) Thank You @Prasad.

 Bapmo said: (Nov 11, 2018) Thanks @Prasad and @Manjunath.N.

 Samruddhia said: (Dec 12, 2018) @All. A leap year calendar repeats itself in 28 years and an Ordinary year Calender repeats itself in 6 or 11 years. Here, we can simply add 2007+6=2013 (which is not there in the option) and 2007+11= 2018 which is the answer.

 Sheetal said: (Dec 28, 2018) In these types of questions, divide last 2digits of the given year from 4, after that if divisible is-- 2 and 3 add 11in it, 1 add 6 in it, 0 add 28 in it. Eg.2007(07/4) divisible is 3 so add 11 in 2007= 2018.

 Manish said: (Jun 14, 2019) 2007 is a general year so either it will repeat after year or 11 year let's see; Year : 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017. Odd day : 1 1 2 1 1 1 2 1 1 1 2 . If the addition is 7 then the calendar will change but 2012 is leap year so general year can't be a leap year so now we go till the 14.

 Pradyumna Kumar Tiwari said: (Jul 1, 2019) 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.

 Pradyumna Kumar Tiwari said: (Jul 1, 2019) The two necessary conditions for the same calendar is; 1) Both years must either be normal or both leap. 2) A number of odd days between two years must be 0.

 Ashwini said: (Aug 16, 2019) How answer is 2018? Please, anyone explain me.

 Fathima Safwana said: (Oct 30, 2019) Is this Holds good for for year where for which year 2009 will be same ?

 Ank said: (Dec 8, 2019) Step1- calculate odd days from given year to that year at which addition is divisible bye 7.

 Mandakranta said: (Feb 24, 2020) 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.

 Meghana said: (Mar 21, 2020) Ordinary year = 1odd day. Leap year = 2odd days. They told to fïnd oud same calendar as 2007, so we should start counting an odd day from 2008, 2008-2odd day 2009-1odd day 2010-1odd day 2011-1odd day 2012-2odd day Add the above odd days. To get same calendar odd days should be multiple of 7. 7 divided by 7 gives reminder 0. Hence 2012 is the answer.

 Kunal said: (May 8, 2020) How the odd days taken here?

 Gireesh Kumar said: (May 17, 2020) @Kunal. See, what is the remainder after dividing the year by 4. If the remainder is 1 add 5 years to the given year. If the remainder is 2 add 11 to the given year. If the remainder is 3 adding 11 to the given year. So in the above question, they gave 2007. After dividing by 4. It gives the remainder as 3. So simply add 11 years to 2007. That will be 2018. Hope it helps you.