Calendar | Calendar Problem Tricks | Calendar Reasoning/Concept/Problems/Questions/Solutions
20Calendar Reasoning | Best Short Tricks in Hindi (2020) | Calendar By Rajesh Choudhary Sir | Best Short Tricks
Introduction
The solar year consists of 3654 days, 5 hrs 48 minutes, 48 seconds. In 47 BC, Julius Ceasar arranged a calendar known as the Julian calendar in which a year was taken as days and in order to get rid of the odd quarter of a day,an extra day was added once in every fourth year and this was called as leap year of Bissextile. Nowadays, the calendar, which is mostly used, is arranged by Pope Gregory XII and known as Gregorian calendar.
In India, number of calendars were being used till recently. In 1952, the Government adopted the National Calender base on Saka era with Chaitra as its first month. In an ordinary year, Chaitra 1 falling on March 22 of Gregorian Calendar and in a leap year it falls on March 21.
Remember
• In an ordinary year,
1 year = 365 days = 52 weeks + 1 day
• In a leap year,
1 year = 366 days = 52 weeks + 2 days
Note : First January 1 A.D. was Monday. So we must count days from Sunday.
• 100 years or century contains 76 ordinary years and 24 leap years.
[76 x 52 days + 76 odd days] + [24 x 52 weeks + 24 x 2 odd days]
= (76 + 24) x 52 weeks + (76 + 48) odd days
= 100x52 weeks + 124 odd days
= 100 x52 weeks + (17 x 7+5) odd days
= (100x52+17) weeks + 5 odd days
100 years contain 5 odd days.
Similarly, 200 years contain 3 odd days.
300 years contain 1 odd days.
400 years contain 0 odd days .
Year whose non-zero numbers are multiple of 4 contains no odd days; like 800, 1200, 1600 etc.
The number of odd days in months
The month with 31 days contains (4x7+3) i.e. 3 odd days and the month with 30 days contains (4x7+2) i.e. 2 odd days.
Note:
February in an ordinary year given no odd days, but in a leap year gives one odd day.
Example: 1. What day of the week was 15th August 1949?
Solution:
15th August 1949 means
1948 complete years + first 7 months of the years
1949 + 15 days of August.
1600 years give no odd day.
300 years give 1 odd day
48 years give {48+12} = 60 =4 odd days.
[For ordinary years 48 odd days and for leap year 1 more day (484) = 12 odd days ; 60=7×8+4]
From 1st January to 15 August 1949 Odd days:
January - 3
February -3
March - 3
April - 2
May - 3
June - 2
July - 3
August - 1
17 3 odd days.
15th August 1949 1 + 4 + 3 = 8 = 1 odd days.
This means that 15th Aug. fell on 1st day. Therefore, the required day was Monday.
Example: 2. How many times does the 28th day of the month occur in 400 consecutive years?
Solution: In 400 consecutive years, there are 97 leap years. Hence, in 400 consecutive years, February has the 29th day 97 times and the remaining eleven months have the 29th days 400 x 100 = 4400 times
The 29th day of the month occurs (4400 + 97) or 4497 times.
Example: 3. Today is 5th February. The day of the week is Tuesday. This is a leap year. What will be the day of the week on this date after 5 years?
Solution: This is a leap year. So, next 3 years will give one odd day each. Then leap year gives 2 odd days and then again next years give 1 odd day.
Therefore (3+2+1)=6 odd days will be there.
Hence the day of the week will be 6 odd days beyond
Tuesday, i.e., it will be Monday.
Example: 4. What day ofthe week was 20th June 1837?
Solution.: 20th June 1837 means 1836 complete years + first 5 months of the year 1837 + 20 days of June.
1600 years give no odd days.
200 years give 3 odd days.
36 years give (36+9) or 3 odd days
1836 years give 6 odd days.
From 1st January to 20th June there are 3 odd days
Odd days : 3
February : 0
March : 3
April : 2
May : 3
June : 6
......
17
Therefore, the total number of odd days = (6+3) or 2 odd days.
This means that the 20th June fell on the 2nd day commencing from Monday. Therefore, the required day was Tuesday.
Example: 5. Prove that the calendar for 1990 will same for 2001 also.
Solution: It is clear that the calendar for 1990 will serve for 2001 if first January of both the years is the same weekdays. For that the number of odd days between 31st December 1989 and 31st December 2000 must be zero. Odd days are as given below .
Year 1900 1991 1992 1993 1994
Odd days 1 1 Leap 1 1
2
1995 1996 1997 1998 1999 2000
1 (Leap) 1 1 1 (Leap)
2 2
Total number of odd days = 14 days = 2 weeks + odd days.