My watch gains 5 seconds in 3 minutes was set right at 7 am. In the afternoon of the same day, when the watch indicates quarter past 4 O'clock, the true time is:

Period upto 5th June, 2002 = (2001 yr + Period from 1.1.2002 to 5.6.2002)
? Odd days in 1600 yr = 0
Odd days in 400 yr = 0
Odd days in 1 ordinary year = 1
Odd days in 2001 year = (0 + 0 + 1) = 1

Number of days from 1.1.2002 to 5.6.2002
January + February + March + April + May + June
= 31 + 28 + 31 + 30 + 31 + 5
= 156 days = 22 weeks + 2 days
= 2 odd days

? Total number of odd days = (1 + 2) = 3
? The required day is Wednesday.

Period upto 17th August, 2010 = (2009 yr + Period from 1.1.2010 to 17.8. 2010)

Counting of odd days
Odd day in 1600 yr = 0
Odd days in 400 yr = 0
9 yr = (2 leap years + 7 ordinary years)
= ( 2 x 2 + 7 x 1) = 1 week + 4 days = 4 odd days

Number of days between 1.1.2010 to 17.8.2010
January + February + March + April + May + June + July + August.
= (31 + 28 + 31 + 30 + 31 + 30 + 31 + 17) days
= 229 days = 32 weeks + 5 odd days

Total number of odd days = (0 + 0 + 4 + 5 ) days
= 9 days = 1 week + 2 odd days
Hence, the required day is Tuesday.

First, we find out the day on 1.03.2005. Period upto 1.3.2005.
= (2004 yr + Period from 1.1.2005 to 1.3.2005)
? Odd day in 1600 yr = 0
Odd day in 400 yr = 0
4 yr = (1 leap year + 3 ordinary years)
= (1 x 2 + 3 x 1) = 5 odd days

Number of days between 1.1.2005 to 1.3.2005,
= January + February + March
= 31 + 28 + 1
= 60 days = (8 weeks + 4 days)
= 4 odd days

Total number of odd days
= (0 + 0 + 5 + 4) = 9
= 1 weak + 2 odd days
? 1.3. 2005 was Tuesday.

So, Saturday will fall on 5.3.2005.
Hence, Saturday lies on 5th, 12th, 19th, and 26th of March 2005.

As per the given above question , we know that
August 15, 1947 = ( 1600 + 300 + 46 ) years + January 1 to August 15th, of 1947
August 15, 1947 = ( 1600 + 300 + 46 ) years + ( 365 - August 16 to December 31 1947 )
1600 + 300 + 46 ) years
August 15, 1947 = ( 1600 + 300 + 46 ) years + ( 365 - 138 ) days
Now , 1600 years give 0 odd day
300 years give 1 odd day
46 years give 1 odd day
Number of odd days = 0 + 1 + 1 (from 11 leap years and 35 ordinary years) + 3 = 5 odd days
Hence , The day was Friday.

We can say that ,
During the interval we have two leap years as 1992 and 1996 and it contains February of both these years.
? The interval has ( 5 + 2 ) = 7 days = 0 odd day.
Hence, January 7, 1997 was also Tuesday.

According to question , we know that
Starting with 2000, count for number of odd days in successive years till the sum is divisible by 7.
2000 + 2001 + 2002 + 2003 + 2004 = 2 + 1 + 1 +1 + 2 = 7 odd days
? Number of odd days up to 2004 = 0 odd day
Hence , Calendar for 2000 will serve for 2005 also.