Calendar Questions

Given year 1856, when divided by 4 leaves a remainder of 0.

NOTE: When remainder is 0, 28 is added to the given year to get the result.

So, 1856 + 28 = 1884

9th May, 2001 = (2000 years + Period from 1.1.2001 to 9.5.2001)

Odd days in 1600 years = 0

Odd days in 400 years = 0

Jan. Feb. March April May 

(31 + 28 + 31 + 30 + 9) = 129 days

129 days = (18 weeks + 3 day)= 3 odd day.

Total number of odd days = (0 + 0 + 3 ) = 3 odd day.

Given day is Wednesday.

Given year, 2013 when divided by 4 leaves a remainder of 1.

NOTE: When the remainder is 1, 11 is substracted from the given year to get the result.

So, 2013 - 11 =2002

31st Dec, 2011 = (2010 years + Period from 1.1.2011 to 31.12.2011)

Odd days in 2000 years = 0

10 years has 2 leap years + 8 ordinary years.

Number of odd days in 10 years ( 2 x 2 + 8) = 5 odd days.

31st,Dec => complete year of 2011 (non-leap year) = 1 odd day.

Total number of odd days = (0 + 5 + 1) = 6.

Given day is saturday.