What was the days of the week on 17th August , 2010?

Period upto 17th July, 1776 = 1775 yr + Period from 1st January 1776 to 17th July 1776

Counting of odd days
In 1600 yr = 0
In 100 yr = 5
75 yr = 18 leap years + 57 ordinary years = (18 x 2 + 57 x 1) odd days
= 93 odd days
= 13 weeks + 2 days
= 2 odd days

? 1775 yr have(0 + 5 + 2) odd days
= 7 odd days = 1 week + 0 odd day
= 0 odd day

Number of days between 1.1.1776 to 17.7.1776
January + February + March + April + May + June + July
31 + 29 + 31 + 30 + 31 + 30 + 17 = 199 days = 28 weeks + 3 odd days

? Total number of odd days = (3 + 0) = 3
Hence, the required day is Wednesday.

Number of odd days between 6th November, 1987 and 4th April, 1988 .
= (30 - 6) days of November 1987+ 31 days of December 1987 + 31 days of January 1988 + 29 days of February
1988 + 31 days March 1988 + 4 days of April 1988
= {(30 - 6 ) + 31 } + (31 + 29 + 31 + 4)
= 24 + 31 + 31 + 29 + 31 + 4 = 150
= 21 weeks + 3 days = 3 odd days
? 6th November, 1987 will be three days backward from Monday.
? Required day is Friday.

6th March 2005 = Monday.
Then, 6th March 2004 = Monday - 1 day = Sunday
[ ? 2004 is a leap year but it does not cross 29th February of 2004, so only 1 is taken as odd day. ]

6th March 2004 = Sunday
7th march 2004 = Monday.

Time from 7 am on Sunday to 7 pm on following Sunday
= 7 days 12 h = 180 h
? Watch gains (5 + 5⁴/₅) min or 54/5 in 180 h.
Now, 54/5 min are gained in 180 h.
? 5 min gained in (180 x 5/54 x 5) h = 83 h 20 min.
= 3 days 11 h and 20 min

? Watch is correct after 3 days 11 h and 20 min after 7 am of Sunday.
? It will be correct at 20 min past 6 pm on Wednesday.

Given that n = 5 and n + 1 = 6
The hands will make right angle at
(5n ± 15) x 12/11 min past 5 .
? (5 x 5 ± 15) x 12/11 min past 5.
? (25 + 15) x 12/11 min past 5 and (25 - 15) x 12/11 min past 5.
? 40 x 12/11 min past 5 and 10 x 12/11 min past 5.
? 4801/11 min past 5 and 120/11 min past 5.
? 43⁷/₁₁ min past 5 and 10¹⁰/₁₁ min past 5.

100 yr have 5 odd days.
? Last day of 1st century of is Friday.

200 yr have (5 x 2) = 1 week + 3 odd day = 3 odd day
? Last day of 2nd century of Wednesday,

300 yr have (5 x 3) = 2 week + 1 odd day = 1 odd day
? Last day of 3rd century is Monday.

400 yr have 0 odd day.
? Last day of 4th century is Sunday.

This cycle is repeated.
? Last day of a century cannot be Tuesday of Thursday of Saturday.

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.

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.

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.