28 May, 2006 = (2005 years + Period from 1.1.2006 to 28.5.2006)
Odd days in 1600 years = 0
Odd days in 400 years = 0
5 years = (4 ordinary years + 1 leap year) = (4 x 1 + 1 x 2) ≡ 6 odd days

Jan.         Feb.       March       April         May
(31     +     28     +     31     +     30     +     28 ) = 148 days

∴ 148 days = (21 weeks + 1 day) ≡ 1 odd day.
Total number of odd days = (0 + 0 + 6 + 1) = 7 ≡ 0 odd day.
Given day is Sunday.

Each day of the week is repeated after 7 days.
So, after 63 days, it will be Monday.
∴ After 61 days, it will be Saturday.

x weeks x days = (7x + x) days = 8x days.

100 years contain 5 odd days.
∴ Last day of 1^st century is Friday.
200 years contain (5 x 2) ≡ 3 odd days.
∴ Last day of 2^nd century is Wednesday.
300 years contain (5 x 3) = 15 ≡ 1 odd day.
∴ Last day of 3^rd century is Monday.
400 years contain 0 odd day.
∴ Last day of 4^th century is Sunday.
This cycle is repeated.
∴ Last day of a century cannot be Tuesday or Thursday or Saturday.

The year 2004 is a leap year. It has 2 odd days.
∴ The day on 8^th Feb, 2004 is 2 days before the day on 8^th Feb, 2005.
Hence, this day is Sunday.