How to find the years which have the same Calendars :
Leap year calendar repeats every 28 years.
Here 28 is distributed as 6 + 11 + 11.
Rules:
a) If given year is at 1st position after Leap year then next repeated calendar year is Given Year + 6.
b) If given year is at 2nd position after Leap year then next repeated calendar year is Given Year + 11.
c) If given year is at 3rd position after Leap year then next repeated calendar year is Given Year + 11.
Now, the given year is 2018
We know that 2016 is a Leap year.
2016 2017 2018 2019 2020
LY 1st 2nd 3rd LY
Here 2018 is at 2 nd position after the Leap year.
According to rule ( b) the calendar of 2018 is repeated for the year is 2018 + 11 = 2029.
Given year is divided by 4, and the quotient gives the number of leap years.
Here, 300%4 = 75.
But, as 100,200 and 300 are not leap years => 75 - 3= 72 leap years.
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
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. .
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.
We will show that the number of odd days between last day of February and last day of October is zero. .
March April May June July Aug. Sept. Oct.
31 + 30 + 31 + 30 + 31 + 31 + 30 + 31
= 241 days = 35 weeks = 0 odd day. ,Number of odd days during this period = 0.
Thus, 1st March of an year will be the same day as 1st November of that year. Hence, the result follows
We already know that the calendar after a leap year repeats again after 28 years.
Here 1988 is a Leap year, then the same calendar will be in the year = 1988 + 28 = 2016.
If yesterday were tomorrow, today is Friday => tomorrow of friday is Saturday.
Hence, it implies that today is Sunday if tomorrow is Saturday.
15th August, 1947 = (1946 years + Period from 1st Jan., 1947 to 15th )
Counting of odd days:
1600 years have 0 odd day. 300 years have 1 odd day.
47 years = (11 leap years + 36 ordinary years)= [(11 x 2) + (36 x 1) ]odd days = 58 odd days = 2 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,
Total number of odd days = (0 + 1 + 2 + 3) odd days = 6 odd days.
Hence, the required day was 'Saturday'.
First,we count the number of odd days for the left over days in the given period.Here,given period is 09-07-2013 to 07-01-2014
Month | Jul | Aug | Sep | Oct | Nov | Dec | Jan |
---|---|---|---|---|---|---|---|
Days | 22 | 31 | 30 | 31 | 30 | 31 | 7 |
Odd Days | 1 | 3 | 2 | 3 | 2 | 3 | 0 |
Therefore, No. of Odd Days = 1 + 3 + 2 + 3 + 2 + 3 + 0 = 14 = 0 odd days
So, given day Sunday + 0 = Saturday is the required result.
Comments
There are no comments.Copyright ©CuriousTab. All rights reserved.