If the period between the two months is divisible by 7, then that two months will have the same calender .
(a). Oct + Nov = 31 + 30 = 61 (not divisible by 7)
(b). Apr + May + Jun + Jul + Aug + Sep + Oct = 30 + 31 + 30 + 31 + 31 + 30 + 31 = 214 (not divisible by 7)
(c). Jun + July + Aug + Sep = 30 + 31 + 31 + 30 = 122 (not divisible by 7)
(d). Apr + May + June = 30 + 31 + 30 = 91 (divisible by 7)
Hence, April and July months will have the same calendar.
We shall find the day on 1st April, 2001.
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 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.
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 6th March, 2005 will be 1 day beyond the day on 6th March, 2004.
Given that, 6th March, 2005 is Monday.
6th March, 2004 is Sunday (1 day before to 6th March, 2005).
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 = 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 = 4 odd days.
Jan. Feb. Mar. Apr. May. Jun. Jul. Aug.
(31 + 28 + 31 + 30 + 31 + 30 + 31 + 15) = 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
x weeks x days = (7x + x) days = 8x days.
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
S0, 8th Dec, 2006 is Friday.
This can be illustrated with an example.
Ex: 1896 is a leap year.The next leap year comes in 1904 (1900 is not a leap year).
First,we count the number of odd days for the left over days in the given period.
Here,given period is 15.8.2012 to 11.6.2013
Aug Sept Oct Nov Dec Jan Feb Mar Apr May Jun
16 30 31 30 31 31 28 31 30 31 11(left days)
2 + 2 + 3 + 2 + 3 + 3 + 0 + 3 + 2 + 3 + 4 (odd days) = 6 odd days
So,given day Thursday + 6 = Wednesday is the required result.
We know that,
Odd days --> days more than complete weeks
Number of odd days in 400/800/1200/1600/2000 years are 0.
Hence, the number of odd days in first 1600 years are 0.
Number of odd days in 300 years = 1
Number of odd days in 49 years = (12 x 2 + 37 x 1) = 61 days = 5 odd days
Total number of odd days in 1949 years = 1 + 5 = 6 odd days
Now look at the year 1950
Jan 26 = 26 days = 3 weeks + 5 days = 5 odd days
Total number of odd days = 6 + 5 = 11 => 4 odd days
Odd days :-
0 = sunday ;
1 = monday ;
2 = tuesday ;
3 = wednesday ;
4 = thursday ;
5 = friday ;
6 = saturday
Therefore, Jan 26th 1950 was Thursday.
First,we count the number of odd days for the left over days in the given period.
Here,given period is 12.2.1986 to 1.1.1987
Feb Mar Apr May June July Aug Sept Oct Nov Dec Jan
16 31 30 31 30 31 31 30 31 30 31 1 (left days)
2 + 3 + 2 + 3 + 2 + 3 + 3 + 2 + 3 + 2 + 3 + 1(odd days) = 1 odd day
So,given day Wednesday + 1 = Thursday is the required result.
Comments
There are no comments.Copyright ©CuriousTab. All rights reserved.