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.
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.
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.
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.
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'.
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