Aptitude Calendar Theory

Formulas:

Odd Days:

We are supposed to find the day of the week on a given date.

For this, we use the concept of 'odd days'.

In a given period, the number of days more than the complete weeks are called odd days.

Leap Year:

(i). Every year divisible by 4 is a leap year, if it is not a century.

(ii). Every 4th century is a leap year and no other century is a leap year.

Note: A leap year has 366 days.

Examples:

Each of the years 1948, 2004, 1676 etc. is a leap year.

Each of the years 400, 800, 1200, 1600, 2000 etc. is a leap year.

None of the years 2001, 2002, 2003, 2005, 1800, 2100 is a leap year.

Ordinary Year:

The year which is not a leap year is called anordinary years. An ordinary year has 365 days.

Counting of Odd Days:

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

$\therefore$ 1 ordinary year has 1 odd day.

2.1 leap year = 366 days = (52 weeks + 2 days)

$\therefore$ 1 leap year has 2 odd days.

3.100 years = 76 ordinary years + 24 leap years

= (76 x 1 + 24 x 2) odd days = 124 odd days.

= (17 weeks + days) 5 odd days.

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 days.

Calendar Problems

Description :

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.

Note:
An ordinary year has one day while a leap year has two 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.

Note: A leap year has 366 days(52 complete weeks + 2 extra days = 366 days).

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.

Type 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 =
(365 – 5)days of the year 1991 +
31 days of January 1992 +
29 days of February 1992 +
4 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.

Exercise:

Type 2:

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

Solution:
Given that,
4th June, 2002 = (Odd days in 2000 years+ 2001 years + period from 1.1.2002 to 4.6.2002)
Odd days in 2000 years=0 Odd days in 2001 year = 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.

Exercise:

Type 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 (Number of odd days in 300 years = (5 x 3) 1 odd day.)+ 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.

Exercise:

Type 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.

Exercise:

Type 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.

Exercise:

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