Aptitude - Calendar - Discussion

When we divide the number of days of a year with 7 it gives the number complete weeks in that year,and the reminder of this gives the number of days which are remaining in the year.these days are called as odd days (i.e.an incomplete week).

One year = number of complete weeks+number of odd days.

I am not getting this odd days. Total number of odd days = (0 + 0 + 6 + 1) = 7 0 odd day.

Can you please explaine me?

2011 year code will be 6. How can you explain this ?

Please explain me clearly. I am totally confused.

I am totally confused in calander question. Then please expain in a easy lang if any one can.

Hi Naveen thank you for your information.

Your method is good I like this method.

Please tell me how to know this method and give me some ideas.

Hi All,

Deena s solution is correct. Its 106 not 16... But try to substitute 106 instead of 16.. We are getting the same answer. I tried with some other questions also,its working.. Good Solution Deena.... Thanks :-)

Deena's solution is correct just take 106 instead of 16 106/4 you will get quotient 26 and remainder 2 leave the remainder and take 26 you will get the right answer.

@NAVEEN. Thanks. !

Dont know how 1600 and 400 have been accounted.
Number of leap yrs in 2000 yrs= 2000/4-15=485 yrs
(4 is included because leap year comes generally once in 4 yrs.
15 is subtracted because some of yrs namely 100,200,300,500,600 700,900,1000,1100,1300,1400,1500,1700,1800,1900 are not leap yrs in 2000 yrs)
Number of odd days in leap yrs= Remainder of 485*2/7 = 4 days
Number of non leap yrs in 2000 yrs= 2000-485 = 1515 yrs
Number of odd days in non leap yrs= Remainder of 1515/7 = 3 Days
Therefore number of odd days in 2000 yrs= Remainder of(4+3)/7= 0
And for next five yrs from 2001 to 2005 and onwards it is been described in solution.