On 31st December, 2005 it was Saturday.
Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days.
On 31st December 2009, it was Thursday.
Thus, on 1st Jan, 2010 it is Friday.
The century divisible by 400 is a leap year.
$\therefore$ The year 700 is not a leap year.
Given that first day of a normal year was Friday
Odd days of the mentioned year = 1 [Since it is an ordinary year]
Hence First day of the next year = Friday + 1 Odd day = Saturday
Therefore, last day of the mentioned year = Friday
The year 2007 is an ordinary year. So, it has 1 odd day.
1st day of the year 2007 was Monday.
1st day of the year 2008 will be 1 day beyond Monday.
Hence, it will be Tuesday.
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 $\equiv$ 6 odd days
Clearly it can be understood from the question that 9 days ago was a Thursday
Number of odd days in 9 days = 2 [As 9-7 = 2, reduced perfect multiple of 7 from total days]
Hence today = Thursday + 2 odd days = Saturday
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.
$\therefore$ 8th Dec, 2006 is Friday.
Given that 1st October is Sunday
Number of days in October = 31
31 days = 3 odd days
As we can reduce multiples of 7 from odd days which will not change anything
Hence 1st November = Sunday + 3 odd days = Wednesday
Given that seventh day of a month is three days earlier than Friday
=> Seventh day is Tuesday
=> 14th is Tuesday
=> 19th is Sunday
59 days = 8 weeks 3 days = 3 odd days
Hence if today is Thursday, After 59 days, it will be = Thursday + 3 odd days
17th June, 1998 = 1997 years + Period from 1.1.1998 to 17.6.1998
Odd days in 1600 years = 0
Odd days in 300 years = $\left(5 \times 3\right)$ $\equiv$ 1
97 years has 24 leap years + 73 ordinary years.
Number of odd days in 97 years $\left( 24 \times 2 + 73\right)$ = 121 = 2 odd days.
Given that 25th August = Thursday
Hence 29th August = Monday
So 22nd,15th and 8th and 1st of August also will be Mondays
Number of Mondays in August = 5
We cannot find out the answer because the number of days of the current month is not given.
Solution:
As it is known, that
Number of days between 1st January 1992 and 1st January 1993 =
(366-1=365)days of the year 1992( as 1992 is completely divisible by 4, hence it is a leap year and that’s why February has 29 days)+
1 days of January 1993
= 365+ 1= 366
= 52 weeks + 2 odd days
Hence, number of odd days = 2
1st January 1993 will be the 2nd day beyond Wednesday.
So, the required day will be Friday.
Given that,
4th June, 2002 = (Odd days in 1600 years+ 1601 years + period from 1st jan, 1602 to 2nd March, 1602)
Odd days in 1600 years=0
Odd days in 1601 year = 1
Number of days according to the months are
January = 31 days
February = 28 days
March = 2 days
Therefore, 31 + 28 + 2=30 = 4 weeks 2 days ≡ 2 odd day
Total number of odd days = 1 + 2 = 3
Therefore, required day is Wednesday.
As known,
Number of odd days upto 30 june 1974 =
(odd days for 1600 years +
odd days for 300 years (Number of odd days in 300 years = (5 x 3) 1 odd day.)+
odd days for 73 years )
= 0 + 1 + ((18 x 2) + (1 x 55) )
= 0 + 1 + 0=1 odd days
odd days of 30 days of june 1974=
Number of days according to the months are
January = 31 days
February = 28 days
March = 31 days
April = 30 days
May = 31 days
June = 30 days
Therefore, 31 + 28 + 31 + 30 + 31 + 30 = 181 = 25 weeks 6 days ≡ 6 odd day
total odd days=1+6=7
Therefore ,30th june 1974 was a sunday.
Sum of odd days from 2003 to 2013 should be zero.
From 2003 to 2013, no of odd days are 1,2,1,1,1,2,1,1,1,2,1 respectively.
Sum of odd days = 1+2+1 + 1 + 1 + 2 + 1 + 1+1+2+1 = 2 week 0 odd days.
So, both dates 1.1.2009 and 1.1.2015 will be on same day, so calendar for the year 2003 will serve for the year 2014.
For this find the day of 1.8.1947
number of odd days for 1600 years=0
number of odd days for 300 years=1
From 1991 to 1946, there are 11 leap years + 35 ordinary years.
So, number of odd days = 11 x 2 + 35 x 1=22+35=57=8weeks+1 odd days=1 odd day
Number of days according to the months are
January = 31 days=3 odd days
February = 28 days=0 odd day
March = 31 days=3 odd days
April = 30 days=2odd days
May = 31 days=3odd days
June = 30 days=2odd days
July =31 days=3odd days
August=1 day
Therefore, 3 + 0+ 3 + 2 + 3 + 2+3+1 =17=2 weeks+3 odd days
Total number of odd days = 1+1+3=5
So, 1.8.1947 will be Friday.
Now, Friday will be on 1.8.1947, 8.8.1947, 15.8.1947, 22.8.1947