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.
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 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.
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).
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.
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.
To determine the day of the week for 8th February 2004, we need to consider the number of days between 8th February 2004 and 8th February 2005. Here's how we can solve this:
The day of the week on 8th February 2004 was Sunday.
Given year is divided by 4, and the quotient gives the number of leap years.
Here, 100%4 = 25.
But, as 100 is not a leap year => 25 - 1= 24 leap years.
First,we count the number of odd days for the left over days in the given period.Here,given period is 18.3.1994 to 25.2.1995
Month | Mar | Apr | May | Jun | Jul | Aug | Sep | Oct | Nov | Dec | Jan | Feb |
---|---|---|---|---|---|---|---|---|---|---|---|---|
Days | 13 | 30 | 31 | 30 | 31 | 31 | 30 | 31 | 30 | 31 | 31 | 25 |
Odd Days | 6 | 2 | 3 | 2 | 3 | 3 | 2 | 3 | 2 | 3 | 3 | 4 |
Therefore, No. of Odd Days = 6 + 2 + 3 + 2 + 3 + 3 + 2 + 3 + 2 + 3 + 3 + 4 = 36 = 1 odd day
So, given day Friday + 1 = Saturday is the required result.
Let us find the day on 1st July, 2004.
2000 years have 0 odd day. 3 ordinary years have 3 odd days.
Jan. Feb. March April May June July
31 + 29 + 31 + 30 + 31 + 30 + 1
= 183 days = (26 weeks + 1 day) = 1 t .
Total number of odd days = (0 + 3 + 1) odd days = 4 odd days. '
:. 1st July 2004 was 'Thursday',-,-
Thus, 1st Monday in July 2004 _as on 5th July.
Hence, during July 2004, Monday fell on 5th, 12th, 19th and 26th. .
NOTE :
Repetition of leap year ===> Add +28 to the Given Year.
Repetition of non leap year
Step 1 : Add +11 to the Given Year. If Result is a leap year, Go to step 2.
Step 2: Add +6 to the Given Year.
Solution :
Given Year is 1993, Which is a non leap year.
Step 1 : Add +11 to the given year (i.e 1993 + 11) = 2004, Which is a leap year.
Step 2 : Add +6 to the given year (i.e 1993 + 6) = 1999
Therfore, The calendar for the year 1993 will be same for the year 1999
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