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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.
January 1st 1992 was a Wednesday. what day of the week was January 1, 1993?
1992 being a leap year it had two odd days show the first day of the year 1993 was today beyond so Wednesday i.e it was Friday.
It was Sunday on Jan 1, 2006. What was the day of the week Jan 1, 2010?
On 31^{st} December, 2005 it was Saturday.
Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days.
∴ On 31^{st} December 2009, it was Thursday.
Thus, on 1^{st} Jan, 2010 it is Friday.
28 May, 2006 i.e. 2005 years and 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 (31) + Feb (28) + March (31) + April (30) + May (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.
Thus, the given day is Sunday.
17^{th} June, 1998 i.e.1997 years + Period from 1.1.1998 to 17.6.1998
Odd days in 1600 years = 0
Odd days in 300 years = 5 x 3 = 1
97 years = 24 leap years + 73 ordinary years.
Number of odd days in 97 years ( 24 x 2 + 73) = 121 = 2 odd days.
Jan (31) + Feb (28) + March (31) + April (30) + May (31) + June (17) = 168 days
Therefore 168 days = 24 weeks = 0 odd day.
Total number of odd days = (0 + 1 + 2 + 0) = 3.
Thus, the given day is Wednesday.
15^{th} August, 2010 = (2009 years + Period 1.1.2010 to 15.8.2010)
Odd days in 1600 years = 0
Odd days in 400 years = 0
9 years = (2 leap years + 7 ordinary years) = (2 x 2 + 7 x 1) = 11 odd days ≡ 4 odd days.
Jan. Feb. March April Mayb June July Aug.
(31 + 28 + 31 + 30 + 31 + 30 + 31 + 15) = 227 days
∴ 227 days = (32 weeks + 3 days) ≡ 3 odd days.
Total number of odd days = (0 + 0 + 4 + 3) = 7 ≡ 0 odd days.
Given day is Sunday.
If 6^{th} March, 2005 is Monday, what was the day of the week on 6^{th} March, 2004?
The year 2004 is a leap year. So, it has 2 odd days.
But, Feb 2004 not included because we are calculating from March 2004 to March 2005. So it has 1 odd day only.
∴ The day on 6^{th} March, 2005 will be 1 day beyond the day on 6^{th} March, 2004.
Given that, 6^{th} March, 2005 is Monday.
∴ 6^{th} March, 2004 is Sunday (1 day before to 6^{th} March, 2005).
We shall find the day on 1^{st} April, 2001.
1^{st} April, 2001 = (2000 years + Period from 1.1.2001 to 1.4.2001)
Odd days in 1600 years = 0
Odd days in 400 years = 0
Jan. Feb. March April
(31 + 28 + 31 + 1) = 91 days ≡ 0 odd days.
Total number of odd days = (0 + 0 + 0) = 0
On 1^{st} April, 2001 it was Sunday.
In April, 2001 Wednesday falls on 4^{th}, 11^{th}, 18^{th} and 25^{th}.
x weeks x days = (7x + x) days = 8x days.
On 8th Feb, 2005 it was Tuesday. What was the day of the week on 8th Feb, 2004?
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.
Count the number of odd days from the year 2007 onwards to get the sum equal to 0 odd day.
Sum = 14 odd days ≡ 0 odd days.
∴ Calendar for the year 2018 will be the same as for the year 2007.
The century divisible by 400 is a leap year.
∴ The year 700 is not a leap year.
On 8th Dec, 2007 Saturday falls. What day of the week was it on 8th Dec, 2006?
The year 2006 is an ordinary year. So, it has 1 odd day.
So, the day on 8th Dec, 2007 will be 1 day beyond the day on 8th Dec, 2006.
But, 8th Dec, 2007 is Saturday.
∴ 8th Dec, 2006 is Friday.
January 1, 2008 is Tuesday. What day of the week lies on Jan 1, 2009?
The year 2008 is a leap year. So, it has 2 odd days.
1^{st} day of the year 2008 is Tuesday (Given)
So, 1^{st} day of the year 2009 is 2 days beyond Tuesday.
Hence, it will be Thursday.
January 1, 2007 was Monday. What day of the week lies on Jan. 1, 2008?
The year 2007 is an ordinary year. So, it has 1 odd day.
1^{st} day of the year 2007 was Monday.
1^{st} day of the year 2008 will be 1 day beyond Monday.
Hence, it will be Tuesday.
22 Apr 2222 i.e. 2221 years + period from 1Jan2222 to 22Apr2222
We know that number of odd days in 400 years = 0
Hence the number of odd days in 2000 years = 0 (Since 2000 is a perfect multiple of 400)
Number of odd days in the period 20012200 = Number of odd days in 200 years = 5 x 2 = 10 = 3
As we can reduce perfect multiples of 7 from odd days without affecting anything
Number of odd days in the period 22012221 = 16 normal years + 5 leap years = 16 x 1 + 5 x 2 = 16 + 10 = 26 = 5 odd days
Number of days from 1Jan2222 to 22 Apr 2222 = 31 (Jan) + 28 (Feb) + 31 (Mar) + 22(Apr) = 112
112 days = 0 odd day
Total number of odd days = (0 + 3 + 5 + 0) = 8 = 1 odd day = Monday
Hence 22 Apr 2222 is Monday
Each day of the week is repeated after 7 days. So, after 63 days, it will be on Monday. After 61 days, it will be on Saturday.
350 days, 350/7 = 50, no odd days, so it will be a Monday.
If today is Saturday, then what day of the week will be on the 338 days from today?
number of odd days in 338 days = 338 / 7 = 248 complete weeks + 2odd days. 2nd day after Saturday is Monday.
Formula : (Date + Month code + No.of years + No.of leap year + Century code)/7
= 43/7 =1
Thus, the day of the week on 24th July 2011will be Sunday.
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