CALENDERS
(i) Odd days
We know that a week contains 7 days counting from monday to Sunday. So, any number of days, which are
more than complete number of a week in a given period are called odd days. For example a period of 10 days contains
3 odd days, 11 days contains 4 odd days, 12 days contains 5 odd days. But period of 14 days contains zero odd day.
Therefore, in finding number of odd days in a given period of time, one has to divide that period by 7. If it is completely
divisible by 7, it contains zero odd day and if it is not divisible by 7 then remaining number of days are the odd days.
(ii) Leap year
Every year which is divisible by 4 is called a leap year. But every century which is divisible by 4 is not a
leap year. Every fourth century is a leap year. For example 400, 800, 1200, 1600..........are all leap years
but centruries like 100, 300, 500, 600........are not leap years.
An ordinary year has 365 days
i.e. (52 weaks + 1 day)
A leap year has 366 days i.e.
(52 weaks + 2 day)
When we divide 365 (an ordinary year) by 7, we get remainder 1, it means that has 1 odd day. Like wise 366 days
(leap year) has 2 odd days. A century has 100 years and every fourth year is a leap year. We can break a century in the leap year as follows :
4, 8, 12, 16, 20..........96.
Now number of terms contained by the above series. 96 = a + (n – 1) d
[Artithmetic progression where a = first term, d = common difference]
96 = 4 + (n – 1)4
92 = (n – 1)4
(n – 1) =92/4 = 23
n = 24
Therefore, a century has 76 ordinary years
100 years = 24 leap year + 76 ordinary year
So, 100 years contain 5 odd days
200 years contain 10 odd days or 3 odd days
300 years contain 15 odd days or 1 odd day
400 years contain 0 odd day.
Like wise years 800, 1200, 1600, 2000......contain zero odd day.
Counting of day in respect of odd day :
Ex.1 Find the day of the week on :
(A) 27th Dec. 1985 (B) 15th Aug. 1947
(C) 12th Jan. 1979
Sol.(A) 27th December 1985 has
(1984 years, 11 months and 27 days).
Now 1600 years have 0 odd day
300 years have 1 odd day
84 year contins
= (21 leap years and 63 ordinary years)
= (21 × 366 + 63 × 365) days
= (7686 + 22995) = 30681 days = 4383 weeks
i.e. 84 years contains 0 odd day.
11 months and 27 days = 361 days
51 weeks + 4 days = 4 odd days.
= 4 + 1 = 5
Therefore,
27 December 1985 has (4 + 1) = 5 odd days.
Now counting sunday as 0 odd day, Tuesday as 2 odd day, and so on. Friday
will have 5 odd day. Therefore, 27th December 1985 will be Friday.
(B) 15th August 1947
(1946 years, 7 months and 15 days)
Now 1600 years have 0 odd day
300 years have 1 odd day
1900 years have 1 odd day
46 years have (11 leap years and 35 ordinary years)
= (11 × 366 + 35 × 365) days = 16801 days
= (2400 weeks + 1 day) = 1 odd day.
7 months and 15 days = 227 days
(32 weeks + 3 days) = 3 odd days
(1946 years + 7 months + 15 days)
have (1 + 1 + 3) = 5 odd days
Which is Friday.
(C) 12th Jan. 1979 = (1978 years + 12 days)
Now, 1600 years have 0 odd days
300 years have 1 odd day
78 years have
(19 leap years + 59 ordinary years)
(19 × 366 + 59 × 365) day = 28489 days
= 4069 weeks + 6 days = 6 odd days
12 days = 5 odd days
Total number of odd days
= (1 + 6 + 5)
= 12 = 5 odd days.
So, the day on 12th January 1979 was Friday.
Ex.2 On what days of July 1776 did sunday fall
Sol . First of all find the day on Ist July 1776.
Ist July 1776 =
(1775 years + 6 months + 1 days)
Now, 1600 years have 0 odd days
100 years have 5 odd day
75 year have 18 leap years and 57 ordianary years
Which have 2 odd days
1775 years have (0 + 5 + 2) = 7 = 0 odd day
Now, 6 months + 1 day
= Jan + Feb + March + April + May+June+1
= (31 + 29* + 31 + 30 + 31 + 30) + 1
= 183 days = 1 odd day
Ist July 1776 will be Monday and hence first Sunday for the month of July
will fall on 7th. Therefore, other Sunday will fall on 14th, 21st and 28th.
Ex.3 What was the day on 26th January 1950, when first Republic Day of India was celebrated
(A) Monday (B) Tuesday
(C) Thurday (D) Friday
Sol. 26th January 1950 menas
(1949 years and 26 days)
1600 years have 0 odd days
300 years have 1 odd day
49 years have
(12 leap years and 37 ordinary years)
Þ (12 × 366 + 37 × 365) days
Þ (4392 + 13505) days
Þ (17897) days = 2556 weeks + 5 days
So, 49 years have 5 odd days
and 26 days have 5 odd days
Total number of odd days
= 0 + 1 + 5 + 5
= 11 days = 4 odd days.
Hence the day on 26th January 1950 was Thursday.
Ex.4 What is the number of odd days in a leap year?
(A) 1 (B) 2 (C) 3 (D) 4
Sol . A leap year has 366 days
Now if we divide 366 by 7 it gives 2 as remainder.
hence number of odd days in 366 days is 2.
Ex.5 Prove that the calender for 1990 will serve for 2001 also.
Sol. The number of odd days between 31st De. 1989 and 31st Dec. 2000. The sum of odd days should be zero.
Odd days are calculated as below :
Sum of odd days = 14 i.e. 0 odd days.
ASSIGNMENT
Q.1 The year next to 1996 will have the same calender as that of the year 1996 :
(A) 2001 (B) 1996
(C) 1997 (D) 1999
Q.2 The year next to 1990 will have the same calendar as that of the year 1990 :
(A) 1995 (B) 1997
(C) 1996 (D) 1992
Q.3 What was the day of the week on 2nd July 1984?
(A) Wednesday (B) Tuesday
(C) Monday (D) Thursday
Q.4 Today is Friday. The day after 63 days will be:
(A) Friday (B) Thursday
(C) Saturday (D) Monday
Q.5 Today is Thursday. The day after 59 days will be :
(A) Sunday (B) Monday
(C) Tuesday (D) Wednesday
Q.6 Today is Wednesday, What will be the day after 94 days ?
(A) Monday (B) Tuesday
(C) Wednesday (D) Sunday
Q.7 On what dates of December, 1984 did Sunday fall ?
(A) 6th, 13th, 20th & 27th
(B) 7th, 14th, 21th & 28th
(C) 2nd, 9th, 16th & 23rd & 30th
(D) 1st, 8th, 15th & 22nd
Q.8 What is the day on Ist January 1901 ?
(A) Monday (B) Wednesday
(C) Sunday (D) Tuesday
Q.9 What is the day on 31st October 1984 ?
(A) Friday (B) Sunday
(C) Wednesday (D) Monday
Q.10 What is the day on 14th March, 1993 ?
(A) Friday (B) Thursday
(C) Sunday (D) Saturday
Q.11 On what dates of August 1980 did Monday fall?
(A) 4th, 11th, 18th & 25th
(B) 3rd, 10th, 17th & 24th
(C) 6th, 13th, 20th & 27th
(D) 9th, 16th, 23th & 30th
ANSWER KEY
1) A 2) C 3) C 4) A 5) A 6) C
7) C 8) D 9) C 10) C 11) A