Please wait ...... Study Card - Calendar Please use your left and right arrow keys to navigate. Use only latest version of browsers like Firefox or Chrome or Opera Introduction Odd Days In Centuries Odd Days - Day Mapping Example 1 - How many odd days are there till the end of year 1900?   Solution: Every 400 years, odd days =0. Hence till 1600 years number of odd days=0. So from 1601 to 1900 it is 300 years. Hence it is 1 odd day (with the help of the table below).   Example 2 - How many odd days are there till the end of year 1910?   Solution: Every 400 years, odd days =0. Hence till 1600 years number of odd days=0. From 1601 to 1900 it is 300 years. Hence it is 1 odd day (with the help of the table below). From 1901 to 1910 we have 2 leap years (1904 and 1908) and 8 ordinary years. Hence odd days = 2*2 + 8*1 = 12 odd days. Total number of odd days = 1 + 12 = 13.  Remainder of 13 divided by 7 is 6. Hence the answer is 6 odd days. Example 3 - How many odd days are there till the end of March, 1911?   Solution: From the previous example we know odd days till the end of 1910 is = 6 odd days. Number of days till the end of March 1911 = Number of days in Jan, Feb and Mar = 31 + 28 (as 1911 is not a leap year)+31 = 90. Odd days in 90 days = remainder of 90 divided by 7 = 6 odd days. Total number of odd days = 6+6 = 12.    Remainder of 12 divided by 7 is 5. Hence the answer is 5 odd days. How to find the day given a date? Example 4 - What will be the day on 15th August, 2010?   Solution: Step 1: Find number of odd days in the passed CENTURY Till 2000, there are 0 odd days. (As 2000 is a multiple of 400 and for every 400 years there are 0 odd days).   Step 2: Find number of odd days in the passed YEAR(S). From 2001 to 2009, there are 2 leap years (2004 and 2008) and 7 ordinary years. Hence odd days = 2*2 + 7*1 = 11 odd days.   Step 3: Find number of odd days in the current year TILL PREVIOUS MONTH END From Jan 2010 to July 2010 there are 31+28 (as 2010 is not a leap year) + 31 + 30 + 31 + 30 + 31 = 212 days. Odd days = remainder of 212 divided by 7 = 2 odd days.   Step 4: Find number of odd days in the current month TILL GIVEN DATE From 01-Aug-2010 to 15-Aug-2010 there are 15 days. Odd days = remainder of 15 divided by 7 = 1 odd day.   Step 5: Add all odd days in above steps. Total odd days = 0+11+2+1= 14   Step 6: Divide by 7 and find the remainder. Map it to the days (0-Sun, 1-Mon,..., 6-Sat) Final odd day count = remainder of 14 divided by 7 = 0.   0 corresponds to Sunday. Hence the 15-Aug-2010 is a Sunday. Example 5 - What will be the day on 17th June, 1998?   Solution: Step 1: Find number of odd days in the passed CENTURY Till 1900, there is 1 odd day. (As 1900 =1600+300. For every 400 years there are 0 odd days and for every 300 years there is one odd day).   Step 2: Find number of odd days in the passed YEAR(S). From 1901 to 1997, there are 24 leap years (1904, 1908, 1912 till and 1996) and 73 ordinary years. Hence odd days = 24*2 + 73*1 = 48+73 = 121 odd days = 2 odd days. (The remainder when 121 divided by 7 is 2)   Step 3: Find number of odd days in the current year TILL PREVIOUS MONTH END From Jan 1998 to May 1998 there are 31+28 (as 1998 is not a leap year) + 31 + 30 + 31 = 151 days. Odd days = remainder of 151 divided by 7 = 4 odd days.   Step 4: Find number of odd days in the current month TILL GIVEN DATE From 01-Jun-1998 to 17-Jun-1998 there are 17 days. Odd days = remainder of 17 divided by 7 = 3 odd days.   Step 5: Add all odd days in above steps. Total odd days = 1+2+4+3=10   Step 6: Divide by 7 and find the remainder. Map it to the days (0-Sun, 1-Mon,..., 6-Sat) Final odd day count = remainder of 10 divided by 7 = 3.   3 correspond to Wednesday. Hence the 17-June-1998 is a Wednesday. Example 6 - It was Sunday on Jan 1, 2006. What was the day on Jan 1, 2010?   Solution: 31-Dec-2005 will be a Saturday (As it was Sunday on Jan 1, 2006)   From 01-Jan-2006 to 31-Dec-2009, there are 1 leap year (2008) and 3 ordinary years. Hence total odd days = 1*2 + 3*1 = 5 odd days.   Adding the day on 01-Jan-2006 total odd days = 5+1 = 6 odd days. Hence Saturday + 6 days = Friday. So 01-Jan-2010 is a Friday. Example 7 - Today is Monday. After 61 days, it will be   In 61 days, number of odd days = remainder of 61 divided by 7 = 5. Hence Monday + 5 days = Saturday Example 8 - If 6th March, 2005 is Monday, what was the day of the week on 6th March, 2004?   From 06-Mar-2004 to 06-Mar-2005, the difference in days = 365 days (As 2005 is an ordinary year, February has only 28 days). Number of odd days = remainder got when 365 is divided by 7 = 1 odd day. Hence Day on 6-Mar-2004 is Monday – 1 odd day (We subtract as date is earlier) = SUNDAY Example 9 - The last day of the century cannot be     From the above data, the last day of 100 years = 5 = Friday,   200 years = 3 = Wednesday, 300 years = 1 = Monday, 400 years = 0 = Sunday. Hence last day of a century cannot be Tuesday or Thursday or Saturday.   Note: From the above fact we can also infer that, the first day of a century CANNOT be Wednesday, Friday or Sunday.