27. CALENDAR
IMPORTANT FACTS AND FORMULA
Under this heading we mainly
deal with finding the day of the week on a particular given date the process of finding it lies on obtaining the
number of odd days.
Odd Days : Number of days more than
the complete number of weeks in a given
Period ., is the number
of odd days during that period.
LeapYear: Every year which is
divisible by 4 is called a leap year.
Thus each one of the years 1992, 1996, 2004, 2008, 2012, etc. is a leap
year. Every 4th century is a leap year but no other century is a leap year.thus
each one of 400, 800, 1200,' 1600, 2000, etc. is a leap year.
None of 1900, 2010, 2020, 2100, etc. is a leap year.
An year which is not a leap year is called an
ordinary year.
(I )An ordinary year has 365 days. (II) A leap
year has 366 days.
Counting of Odd
Days:
i)1 ordinary year = 365 days = (52 weeks + 1 day).
:. An ordinary year has 1 odd day.
ii)1 leap year = 366 days = (52 weeks + 2 days).
:. A leap year has 2 odd days.
_ iii)100 years = 76 ordinary years +
24 leap years
= [(76 x 52) weeks + 76 days) + [(24 x 52) weeks + 48 days]
= 5200 weeks + 124 days = (5217 weeks + 5 days).
:. 100 years contain 5 odd
days.
200 years contain 10 and therefore 3 odd days.
300 years contain 15 and therefore 1 odd day.
400 years contain (20 + 1) and therefore 0 odd day.
Similarly, each one of 800,
1200, 1600, 2000, etc. contains 0 odd days.
Remark: (7n + m) odd days,
where m < 7 is equivalent to m odd days.
Thus, 8 odd days ≡ 1 odd day etc.
No of odd days
|
0
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1
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2
|
3
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4
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5
|
6
|
Day
|
Sun.
|
Mon.
|
Tues.
|
Wed.
|
Thur.
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Fri.
|
Sat.
|
SOLVED EXAMPLES
Ex: 1.Wbat was the day of the week on, 16th
July, 1776?
Sol: 16th July,
1776 = (1775 years + Period from 1st Jan., 1776 to 16th July, 1776)
Counting of odd days :
1600 years have 0 odd day. 100 years have 5
odd days.
75 years = (18 leap years + 57 ordinary
years)
= [(18 x 2) + (57 x 1)] odd days = 93 odd
days
= (13 weeks + 2 days) = 2 odd days.
.. 1775 years have (0 + 5 + 2) odd days = 7
odd days = 0 odd day.
Jan.
Feb. March April
May June July
31 + 29 +
31 + 30
+ 31 +
30 +16 = 198days
= (28 weeks + 2 days) =2days
:. . Total number of odd days = (0 + 2) =
2. Required day was 'Tuesday'.
Ex. 2. What was the day of the week on 16th
August, 1947?
Sol. 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. March April May June
July 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'.
Ex. 3. What was the day of the week on 16th
April, 2000 ?
Sol. 16th April, 2000 = (1999 years + Period
from 1st Jan., 2000 to 16thA'
Counting of odd days:
1600 years have 0 odd day. 300 years have 1
odd day.
99 years = (24 leap years + 75 ordinary
years)
= [(24 x 2) + (75 x 1)] odd days = 123 odd
days
= (17 weeks + 4 days) = 4 odd days.
Jan. Feb. March April
31 + 29 + 31 + 16 = 107 days =
(15 weeks + 2 days) = 2 odd,
Total number of odd days = (0 + 1 + 4 + 2)
odd days = 7 odd days = 0 odd day. Hence, the required day was 'Sunday'.
Ex. 4. On what dates of Jull.2004 did
Monday fall?
Sol .
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. .
Ex. 5. Prove that the calendar for the year
2008 will serve for the year 20ll
Sol. In order that the calendar for the year 2003 and
2014 be the same, 1st January
of both the years must be on the same
day of the week.
For this, the number of odd days between
31st Dec., 2002 and 31st Dec.,2013 must be the same.
We know that an ordinary year has 1 odd
day and a leap year has 2 odd During this period, there are 3 leap years,
namely 2004, 2008 and 2012 and 8 ordinary years.
Total number of odd days = (6 + 8) days = 0
odd day.
Hence, the calendar for 2003 will serve for
the year 2014.
Ex. 6. Prove that any date in March of a
year is the same day of the week corresponding date in November that year.
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 = O.
Thus, 1st March of an year will be the same
day as 1st November of that year. Hence, the result follows.
