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.
"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 = (5 x 3) 1
97 years has 24 leap years + 73 ordinary years.
Number of odd days in 97 years ( 24 x 2 + 73) = 121 = 2 odd days.
Jan. Feb. March April May June
(31 + 28 + 31 + 30 + 31 + 17) = 168 days
168 days = 24 weeks = 0 odd day.
Total number of odd days = (0 + 1 + 2 + 0) = 3.
Given day is Wednesday.
"
"15th 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 May 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.
"
days though may not be in order. Gamma is as many days old to Alpha as Beta is
younger to Epsilon. Delta is two days older then Epsilon. Gammas Birthday is on
Wednesday. Tell whose birthday is when.
Alpha, Beta , gamma, delta and epsilon are friends and have birthdays on consecutive
days though may not be in order. Gamma is as many days old to Alpha as Beta is
younger to Epsilon. Delta is two days older then Epsilon. Gammas Birthday is on
Wednesday. Tell whose birthday is when.
Alpha: Friday
Beta : Saturday
Gamma: Wednesday
Delta: Tuesday
Epsilon: Thursday
Given that 1.12.91 is the first Sunday
Hence we can assume that 3.12.91 is the first Tuesday
If we add 7 days to 3.12.91, we will get second Tuesday
If we add 14 days to 3.12.91, we will get third Tuesday
If we add 21 days to 3.12.91, we will get fourth Tuesday
=> fourth Tuesday = [3.12.91 + 21 days ]= 24.12.91
For a year to have the same calendar with 1990 ,total odd days from 1990 should be 0.
Take the year 1992 from the given choices.
Total odd days in the period 1990-1991= 2 normal years
≡ 2 x 1 = 2 odd days
Take the year 1995 from the given choices.
Number of odd days in the period 1990-1994 = 4 normal years + 1 leap year
≡ 4 x 1 + 1 x 2 = 6 odd days
Take the year 1996 from the given choices.
Number of odd days in the period 1990-1995= 5 normal years + 1 leap year
≡ 5 x 1 + 1 x 2 = 7 odd days ≡ 0 odd days
As we can reduce multiples of 7 from odd days which will not change anything
Though number of odd days in the period 1990-1995 is 0, there is a catch here.
1990 is not a leap year whereas 1996 is a leap year.
Hence calendar for 1990 and 1996 will never be the same.
Take the year 2001 from the given choices.
Number of odd days in the period 1990-2000= 8 normal years + 3 leap years
≡ 8 x 1 + 3 x 2 = 14 odd days ≡ 0 odd days
Also, both 1990 and 2001 are normal years.
Hence 1990 will have the same calendar as that of 2001
1 Jan 1901 = (1900 years + 1st Jan 1901)
We know that number of odd days in 400 years = 0
Hence the number of odd days in 1600 years = 0 [Since 1600 is a perfect multiple of 400]
Number of odd days in the period 1601-1900
= Number of odd days in 300 years
= 5 x 3 = 15 = 1
[As we can reduce perfect multiples of 7 from odd days without affecting anything]
1st Jan 1901 = 1 odd day
Total number of odd days = (0 + 1 + 1) = 2
2 odd days = Tuesday
Hence 1 January 1901 is Tuesday.
100 years contain 5 odd days.
$\therefore$ Last day of 1st century is Friday.
200 years contain 5 x 2 $\equiv$ 3 odd days.
$\therefore$ Last day of 2nd century is Wednesday.
300 years contain 5 x 3 = 15 $\equiv$ 1 odd day.
$\therefore$ Last day of 3rd century is Monday.
400 years contain 0 odd day.
$\therefore$ Last day of 4th century is Sunday.
This cycle is repeated.
$\therefore$ Last day of a century cannot be Tuesday or Thursday or Saturday.
Given that January 1, 2004 was Thursday.
Odd days in 2004 = 2 [because 2004 is a leap year]
Also note that we have taken the complete year 2004 because we need to find out the odd days from 01-Jan-2004 to 31-Dec-2004,
that is the whole year 2004
Hence January 1, 2005 = Thursday + 2 odd days = Saturday
1st 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 $\equiv$ 0 odd days.
Total number of odd days = (0 + 0 + 0) = 0
On 1st April, 2001 it was Sunday.
In April, 2001 Wednesday falls on 4th, 11th, 18th and 25th.
Mentioned month begins on a Saturday and has 30 days
Sundays = 2nd, 9th, 16th, 23rd, 30th
=> Total Sundays = 5
Every second Saturday is holiday.
1 second Saturday in every month
Total days in the month = 30
Total working days = 30 - [5 + 1] = 24
22 Apr 2222 = [2221 years + period from 1-Jan-2222 to 22-Apr-2222]
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 2001-2200
= 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 2201-2221
= 16 normal years + 5 leap years
= 16 x 1 + 5 x 2 = 16 + 10 = 26 = 5 odd days
Number of days from 1-Jan-2222 to 22 Apr 2222
= 31 Jan + 28 Feb + 31 Mar + 22Apr = 112
112 days = 0 odd day
Total number of odd days = 0 + 3 + 5 + 0 = 8 = 1 odd day
1 odd days = Monday
Hence 22 Apr 2222 is Monday.