Calendar deals with odd days, leap year, ordinary year, counting of odd days, and day of the week related to odd days. To find the day of the week on a given date concept of odd days is used.

Extra days, apart from the complete weeks in given periods are called odd days.

**Leap year**:

A leap year is divisible by 4 except for a century. For a century to be a leap year, it must be divisible by 400.

Examples:

- Years like 1988, 2008 are leap year (divisible by 4).

Centuries like 2000, 2400 are leap year ( divisible by 400).

Years like 1999, 2003 are not leap year (not divisible by 4).

Centuries like 1700, 1800 are not leap year ( not divisible by 400).

In a century, there is 76 ordinary year and 24 leap year.

**Ordinary year**:

Ordinary year is other than leap years. A ordinary year has 365 days.

**Counting of odd days**:

(i) 1 ordinary year = 365 days = (52 weeks + 1 day).

An ordinary year has one 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 1 + 24 x 2) odd days = 124 odd days.

= 17 weeks + 5 days ≡ 5 odd days.

Therefore,

Number of odd days in 100 years = 5.

Number of odd days in 200 years = 5 x 2 ≡ 3 odd days.

Number of odd days in 300 years = 5 x 3 ≡ 1 odd day.

Number of odd days in 400 years = 5 x 4 + 1 ≡ 0 odd day.

Similarly, each one of 800 years, 1200 years, 1600 years, 2000 years etc. has 0 odd day.

**Day of the week related to odd days**:

Let the number of days be 0, 1, 2, 3, 4, 5 , 6 and their days are Sunday, Monday, Tuesday, Wednesday, Thursday, Friday and Saturday respectively.

**Examples**:

1. What was the day of the week on 16th july, 17776?

**Solution**:

- 16th july, 1776 = (1775 years + Period from 1.1.1776 to 16.7.1776)

Counting of ood days:

Number of odd days in 1600 years = 0

Number of odd days in 100 years = 5

75 years = 18 leap years + 57 ordinary years

= (18 x 2 + 57 x 1) odd days = 93 odd days

= (13 weeks + 2 days) \(\)\equiv\(\) 2 odd days

∴ 1775 years have = (0 + 5 + 2) odd days = 7 odd days \(\)\equiv\(\) 0 odd days.

31 (Jan) + 29 (Feb) + 31 (March) + 30 (April) + 31 (May) + 30(June) + 16(July) = 198 days

198 days = (28 weeks + 2 days) \(\)\equiv\(\) 2 odd days.

∴ Total number of odd days = (0 + 2) = 2.

Hence, the required day is Tuesday.

2. What was the day of the week on `15th August, 1947?

**Solution**:

- 15th August, 1947 = (1946 years + Period from 1.1.1947 tO 15.8.1947)

Odd days in 1600 years = 0

Odd days in 300 years = (5 x 3) = 15 \(\)\equiv\(\) 1

46 years = (11 leap years + 35 ordinary years)

= (11 x 2 + 35 x 1) odd days = 57 odd days

= (8 weeks + 1 day) = 1 odd day.

∴ Odd days in 1946 years = (0 + 1 + 1) = 2.

31 (Jan) + 28 (Feb) + 31 (March) + 30 (April) + 31 (May) + 30 (June) + 31 (July) + 15 (Aug) = 227 days

227 days = (32 weeks + 3 days) \(\)\equiv\(\) 3 odd days.

Total number of odd days = (2 + 3) = 5.

Hence, the required day is Friday.

3. What was the day of the week on `4th August, 2002?

**Solution**:

- 4th June, 2002 = (2001 years + Period from 1.1.2002 to 4.6.2002)

Odd days in 1600 years = 0

Odd days in 400 years = 0

Odd days in 1 ordinary year = 1

Odd days i 2001 years = (0 + 0 + 1) = 1

31 (Jan) + 28 (Feb) + 31 (March) + 30 (April) + 31 (May) + 4 (June) = 155 days

= 22 weeks + 1 day \(\)\equiv\(\) 1 odd day

Total number of odd days = (1 + 1) = 2

∴ Required days is Tuesday.

- As it is known, that

Number of days between 5th January 1991 and 4th March 1992 = (265 – 5)days of the year 1991 + 31 days of January 1992 + 29 days of February 1992 + 1 day of March 1992 (as 1992 is completely divisible by 4, hence it is a leap year and that’s why February has 29 days).

= 360 + 31 + 29 + 4 = 424

= 60 weeks + 4 days

Hence, number of odd days = 4

4th March 1992 will be the 4th day beyond Saturday.

So, the required day will be Wednesday.

**2. What was the day of the week on 4th June, 2002?**

**Solution**:

- Given that,

4th June, 2002 = (2001 years + period from 1.1.2002 to 4.6.2002)

Odd days in 1600 years = 0

Odd days in 400 years = 0

Odd days in 1 ordinary year = 1

Odd days in 2001 year = (0 + 0 + 1) = 1

Number of days according to the months are

January = 31 days

February = 28 days

March = 31 days

April = 30 days

May = 31 days

June = 4 days

Therefore, 31 + 28 + 31 + 30 + 31 + 4 = 155 = 22 weeks 1 days ≡ 1 odd day

Total number of odd days = 1 + 1 = 2

Therefore, required day is Tuesday.

**3. Find the day of the week on 26th January 1950?**

**Solution**:

- As known,

Number of odd days upto 26th January 1950 = odd days for 1600 years + odd days for 300 years + odd days for 49 years + odd days of 26 days of January 1950.

= 0 + 1 + (12 x 2 + 37 ) + 5

= 0 + 1 + 61 + 5

= 67 days

= 9 weeks + 4 days

= 4 odd days

Thus, it was Thursday on 26th January 1950.

**4. Prove that the calendar for the year 2009 will serve for the year 2015?**

**Solution**:

- Sum of odd days from 2009 to 2014 should be zero.

From 2009 to 2014, no of odd days are 1, 1, 1, 2, 1, 1 respectively.

Sum of odd days = 1 + 1 + 1 + 2 + 1 + 1 = 1 week 0 odd days.

So, both dates 1.1.2009 and 1.1.2015 will be on same day, so calendar for the year 2009 will serve for the year 2015.

**5. On what date of Feb. 2007 did Saturday fall?**

**Solution**:

- For this find the day of 1.2.2007

1600 + 400 years has 0 odd days.

From 2001 to 2006, there are 1 leap years + 5 ordinary years.

So, number of odd days = 1 x 2 + 5 x 1 = 2 + 5 = 7 = 1 week = 0 odd day.

Now, from 1.1.2007 to 1.2.2007 number of days = 32 = 4 weeks + 4 odd days = 4 odd days.

Total number of odd days = 4,

So, 1.2.2007 will be Thursday.

Now, Saturday will be on 3.2.2007.