# Ch.19 :- Calender : Introduction & Formula

**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.

**Leap year :** The 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 4th century is a leap year. For example 400, 800, 1200, 1600,……. are all leap years but centuries like 100, 300, 500, 600,……. are not leap years.

An ordinary year has 365 days i.e., (52 weeks + 1 day)

A leap year has 366 days i.e., (52 weeks + 2 days)

When we divide 365 (an ordinary year) by 7, we get remainder as 1. It means that it has 1 odd day. Like-wise, 366 days (leap year) has 2 odd days.

A century has 100 years and every 4th year is a leap year. We can break a century in the leap years as follows:

4,8, 12, 16,20,…….. ,96.

Now, number of terms contained by the above series is given by

96 = a + (n – 1) d

[Arithmetic progression, where a = 1st term, d = common difference]
=> 96 = 4 + (n-1) 4

or 92 = (n-1) 4

or (n-1) =^{92}⁄_{4}

or n = 24

Therefore, a century has 76 ordinary years 100 years

= 24 leap years + 76 ordinary years

= (24 x 366 + 76 x 365) days

= 36524 days =^{36524}⁄_{7} weeks

= 5217 weeks + 5 days

= 5 odd days.

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.

**Ex. 1. Find the day of the week on **

(a) 27th Dec 1985

(b) 15th Aug 1947

(c) 12th Jan 1979

**Sol. (a)** 27th Dee 1985 has (1984 years, 11 months and 27 days).

Now, 1600 years have 0 odd day.300 years have 1 odd day 84 years contains

= (21 leap years and 63 ordinary years)

= (21 x 366 + 63 x 365) days

= (7686 + 22995) = 30681 days =^{30681}⁄_{7}

= 4383 weeks.

i.e., 84 years contain 0 odd day 11 months and 27 days

= 361 days

= 51 weeks + 4 days

= 4 odd days.

= 4 + 1 = 5

Therefore, 27th Dee 1985 has (6 + 1) = 7 odd days.

Now counting Sunday as 0 odd day, Tuesday as 2 odd days and so on, Friday will have 5 odd days. Therefore, 27th Dee 1985 will be Friday.

**(b)** 15th Aug 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 x 366 + 35 x 365)days = 16801 days

= (2400 week)

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 day 300 years have] odd days.

= (19 x 366 + 59 x 365) days = 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 Jan 1979 was Friday.

**Ex. 2. On what days of July, 1776 did Sunday fall ?**

**Sol.**

First of all find the day on 1st July, 1976.

1st July, 1776 = (1775 years + 6 months + 1 day)

Now, 1600 years have 0 odd day

100 years have 5 odd days

75 years have 18 leap years and 57 ordinary years which have 2 odd days.

1775 years have (0 + 5 + 2) = 7 = 0 odd day

Now, 6 months + 1 day

= Jan + Feb + Mar + Apr + May + June + 1

= (31 + 29″ + 31 + 30 + 31 + 30) + 1

= 183 days

= 1 odd day

∵ 1st July, 1776 win be Monday and hence 1st Sunday for the month of Jury”Will fall on 7th. Therefore, other Sundays will fall on 14th, 21 Sf and 28th.

**Ex. 3. What was the day on 26th Jan 1950, when 1st Republic Day of India was celebrated?**

(a) Monday

(b) Tuesday

(c) Thursday

(d) Friday

**Sol.** 26th Jan, 1950 means

(1949 years and 26 days)

1600 years have 0 odd day

* Because, the year 1776 is a leap year.

300 years have 1 odd day

49 years have (12 leap years and 37 ordinary years)

=> (12 x 366 + 37 x 365)

=> days -(4392 + 13505) days

=> (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 Jan 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 calendars for 1990 will serve for 2001 also .**

**Sol.** There are two ways of proving it. One find the day on 1-Jan, 1990 and 1st Jan, 2001. Both the days should be identical. 2nd find the number of odd days between 31 Dec 1989 and 31St Dee 2000. The sum of odd day

should be zero.

Odd days are calculated as below: