 # Calendar Problems

#### Chapter 23 5 Steps - 3 Clicks

# Calendar Problems

### Introduction

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.

### Concepts

Odd days:
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.

### Samples

1. If 5th January 1991 was Saturday, what day of the week was it on 4th march 1992?

Solution:

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.