It was Saturday on 31^st December 2005.

Odd days from the year 2006 to 2009 = (1+1+2+1) = 5 days

Therefore, it was Thursday on 31^st December 2009, and the next day, 1^st January 2010 will be Friday.

28 May 2006

= (2005 + time from 1^st Jan 2006 to 28^th May 2006)

Number of odd days in 1600 years =0

Number of odd days in 400 years =0

Calculation for 4 years = (4 ordinary years + 1 leap year)

= (4 x 1 + 1x 2)

= 6 odd days

Also, total days =

Jan = 31

Feb = 28

March = 31

April = 30

May= 28

Sum of all these days = 148

Number of add days in these 148 days = (21 weeks + 1 day) = 1 odd day

Total odd days = 0+0+6+7 = 7 (which leaves no odd day)

Therefore, the required day is Sunday.

It takes seven days to repeat each day in a week.

Therefore, after 63 days, it will be on Monday.

And after 61 days, it will be on Saturday.

2004 is a leap year and has 2 odd days

The month of Feb 2004 will not be calculated and hence, it has only 1 odd day.

Therefore, the day of 6^th March 2005 will be one day after the day on 6^th March 2004.

It is also given that 6^th March 2005 is a Monday.

Hence, 6^th March 2004 is a Sunday.

Determining the day on 1^st April 2001 = (2000 + time from 1.1.2001 to 1.4.2001)

Number of odd days in 1600 years = 0

Number of odd days in 400 years = 0

Calculation for Jan, Feb, March, and April= 31 +28 + 31 +1 = 91 days and 0 odd days

Total odd days = (0+0+0) =0

Day on 1^st April 2001 = Sunday

Therefore, dates on which Wednesday will fall = 4, 11, 18, and 25 as counted by the interval of 7 days each.

p weeks and p days = (7p + p) days

= 8p days

Odd days in 100 years = 5

Therefore, we can say that the last day of the first century is Friday.

Odd days in 200 years = (2 x 5) Ξ 3 odd days

Therefore, the last day of the second century is Wednesday

Odd days in 300 years = (5 x 3) = 15  Ξ 1 odd days

Therefore, we can say that the last day of the third century is Monday.

Odd days in 400 years = 0

Therefore, we can say that the last day of the 3^rd centaury is Sunday.

The above cycle is repeated.

We can say that a century’s last day cannot be Tuesday, Thursday, or Saturday.

2004 is a leap year and hence, number of odd days = 2

Therefore, the day of 8^th Feb 2004 will be 2 days before the day on 8^th Feb 2005.

Therefore, the day is Saturday.

The number which is divisible by 400 is a leap year.

Therefore, the year 300 is not a leap year.

2006 is an ordinary year. Hence, the number of odd days = 1.

Also, the day of 7^th December 2007 will be one day after the day on 7^th December 2006.

Day on 7^th December 2007= Friday.

Hence, 8^th December 2006 = Thursday.

2008 is a leap year and has 2 odd days.

The first day is Tuesday, as it is already provided in the question.

Therefore, the first day of year 2009 will be two days beyond Tuesday.

Hence, January 1, 2009, will be Thursday.

It is an ordinary year for 2007 with 1 odd day.

Day of the week for first day of the year = Monday

Day one of the year 2008 = 1 day beyond Monday.

Therefore, the day of week = Tuesday.