Years , Weeks & Days

Introduction

a) Odd Days:

The days more than the complete number of weeks in a given period are called Odd Days

- In any year there are 52 complete weeks.

b) Leap Year :

Below criteria must be considered to identify leap years::

(i) It should be divisible by 4

(ii) And if it is divisible by 100, it should also be divisible by 400.

Note:

- A leap year has 366 days i.e. 52 weeks and 2 odd days.

- Every 4th century is a leap year and no other century is a leap year

Examples:

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

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

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

c) Ordinary Year

The year which is not a leap year is called an Ordinary Years.

An ordinary year has 365 days i.e. 52 weeks and 1 odd days.

d) Counting of Odd Days:

In order to count number of odd days , take note of below points :

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

So 1 ordinary year has 1 odd day.

ii. 1 leap year = 366 days = (52 weeks + 2 days)

1 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 + days) 5 odd days.

Number of odd days in 100 years = 5.

Number of odd days in 200 years = (5 x 2)(mod 7) 3 odd days.

Number of odd days in 300 years = (5 x 3) (mod 7) 1 odd day.

Number of odd days in 400 years = (5 x 4 + 1) (mod 7) 0 odd day.

- Since 400 years is a leap year, it has 1 extra odd days than Ordinary Year.

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

Day of the Week Related to Odd Days:

Month of a Year Related to Odd Days:

i.e. 3 odd days for every month with 31 days,

2 odd days for every month with 30 days

and 0/1 odd days for February.

Important Tips:

The last day of a century must be Monday, Wednesday , Friday or Sunday.

An ordinary year always begins and ends on the same day of the week.

Number of odd days of 1900 years = 1 odd day. Remember this because in most of the questions the year will be some year after 1900

Note 1 : To find the day of the week on a particular date when reference day is given:

Step 1: Find the net number of odd days for the period between the reference date and the given date. (Exclude the reference day but count the given date for counting the number of net odd days)

Step 2:The day of the week on a particular date is equal to the number of net odd days ahead of the reference day (if the reference day was before this date) or behind the reference day (if this date was behind the reference day)

Example 1: 11 January ,1997 was a Sunday . What day of the week was on 7 January , 2000?

Here we need to find the day of 7 January 2000 and reference day for 11 January 1997 is given as Sunday.

So we need to find the Total Number of days between 11 January 1997 and 7 January 2000

= (365 - 11) in 1997 + (365 days in 1998) + (365 days in 1999) + (7 days in 2000)

= (50 weeks + 4 odd days) + (52 weeks + 1 odd day) + (52 weeks + 1 odd day) + (7 odd days)

= 13 odd days = 1 week + 6 odd days

Hence , 7 January 2000 would be 6 days ahead of Sunday (11 January 1997) i.e. it was on Saturday.

Note 2 : To find the day of the week on a particular date when no reference day is given:

Step 1: Count the net number of odd days on the given date

Step 2: Write: -

Sunday for 0 odd days

Monday for 1 odd day

Tuesday for 2 odd days

. . .

. . .

Saturday for 6 odd days

Example 2: What day of the week was on 17 June 1999 ?

In order to find the day of the week on a particular date when no reference day is given , follow below steps :

Step 1 : Find the number of odd days till the previous century of the year in the provided date .

Here given previous century is 1900 . So we need to find the odd days of 1900 years.

1900 years = 1600 years + 300 years = 0 odd days + 1 odd day = 1 odd day

Step 2 : Find the number of odd days from the first year after that century till the previous year of the given year.

Here given year is 1999 i.e. 98 years after century (1900) and till previous year (1999)

So we need to find the odd days of 98 years from 1901 to 1998

98 years have 24 leap years + 74 ordinary years

(Note: We know a leap year occurs after every 4 year .

So to obtain number of leap years in 98 years, divide 98 by 4 which will give 24 leap year)

So Odd days in 98 years = (24 * 2) + (74 * 1) = 122 days = 17 weeks + 3 odd days

(Note: Leap year has 2 odd days and non - leap year has 1 odd day)

So 98 years has 3 odd days.

Step 3 : Find the number of odd days of the completed month of the year in the given date.

Here given completed month is 5 (from January to May)

So 1 January 1999 to 31 May 1999 has odd days

= (3 (Jan) + 0 (Feb since it is non leap year) + 3 (March ) + 2 (April) + 3 (May))

= 11 days = 1 week + 4 day = 4 odd days

Step 4 : Find the number of odd days of the day in the given date.

So we need to find the number of odd days from 1 June to 5 June .

= 5 odd days

Step 5 : Add the result of Step 1 , Step 2 , Step 3 and Step 4 to find the number of odd days corresponding to the given date.

Total No of odd days = 1 + 3 + 4 + 5 = 13 days = 1 week + 6 odd days

Hence 5 June 1999 was Saturday

Example 3: What day of the week was on 15 April 2014 ?

In order to find the day of the week on a particular date when no reference day is given , follow below steps :

Step 1 : Find the number of odd days till the previous century of the year in the provided date .

Here given previous century is 2000 . So we need to find the odd days of 2000 years

= 0 odd days

Step 2 : Find the number of odd days from the first year after that century till the previous year of the given year.

Here given year is 2014 i.e. 13 years after century (2000) and till previous year (2014)

So we need to find the odd days of 13 years from 2001 to 2013

13 years have 3 leap years + 10 ordinary years

(Note: We know a leap year occurs after every 4 year . So to obtain number of leap years in 13 years, divide 13 by 4 which will give 3 leap year)

So Odd days in 13 years = (3 * 2) + (10 * 1)

= 16 days = 2 weeks + 2 odd days

(Note: Leap year has 2 odd days and non – leap year has 1 odd day)

So 13 years has 2 odd days.

Step 3 : Find the number of odd days of the completed month of the year in the given date.

Here given completed month is 3 (from January to March )

So 1 January 2014 to 31 March 2014 has odd days

= (3 (Jan) + 0 (Feb since it is non leap year) + 3 (March ) )

= 6 days = 6 odd days

Step 4 : Find the number of odd days of the day in the given date.

So we need to find the number of odd days from 1 April to 15 April.

2 week + 1 day = 1 odd days

Step 5 : Add the result of Step 1 , Step 2 , Step 3 and Step 4 to find the number of odd days corresponding to the given date.

Total No of odd days = 0 + 2 + 6 + 1

= 9 days = 1 week + 2 odd days

Hence 15 April 2014 was Tuesday

Conditions for calendars of two different years to be same:

1. Both the years must be of the same type (leap or non-leap year). i.e., either both must be leap years or both must be non-leap years.

2. The net number of odd days for the period between these two years should be zero. That is, 1st January of both the years must fall on the same day of the week.

Example 4: The calendar for the year 2015 will be same for the year?

A calendar repeats itself if the number of days between those two years is a multiple of 7.

That is, count the number of odd days from the year 2015 onwards to get the sum equal to 0 odd days and the next year will be our required year.

Sum = 14 odd days 0 odd days.

Thus, by completing the year 2025, there are zero odd days with reference to the year 2015.

The next year is 2026. 2015 and 2026 are both non-leap years.

Hence, Calendar for the year 2026 will be the same as for the year 2015.

Example 5: Calendar for 2000 will be same as the calendar for?

Let's start with the year 2000 to count for number of odd days in successive years till the sum is divisible by 7.

So, Number of odd days up to 2004 = 0

Therefore, Calendar for year 2000 will serve also for 2005

SOLVED EXAMPLES:

Example 6: Which of the below day cannot be last day of the century?

No of odd days in first century (i.e. 100 years) = 5

Therefore: Last day of the first century is Friday

No of odd days in two centuries = 3

Therefore: Last day of the second century is Wednesday

No of odd days in third century = 1

Therefore: Last day of the third century is Monday

No of odd days in fourth century = 0

Therefore: Last day of the fourth century is Sunday

Since the order is continually kept in successive cycles, the last day of a century cannot be Tuesday, Thursday or Saturday.

So, the last day of a century should be Sunday, Monday, Wednesday or Friday.

Therefore, the first day of the century must be either Monday, Tuesday, Thursday or Saturday.

Example 7: In July 2004 on what date did Monday fall?

Let us find the day on 1st July 2004.

2000 years have 0 odd days.

3 ordinary years have 3 odd days.

Jan. (31) + Feb. (29) + March (31) + April (30) + May (31) + June (30) + July (1)

= 183 days = (26 weeks + 1 day) = 1 odd day

Total number of odd days = (0 + 3 + 1) odd days = 4 odd days.

So 1st July 2004 was "Thursday"

Thus, 1st Monday in July 2004 is on 5th July.

Hence, during July 2004, Monday fell on 5th, 12th, 19th and 26th.

Example 8: 1 February 1984 was a Wednesday just like the 29 February 1984. When will the calendar show another February with a similar situation?

The month of February 1984 had 5 Wednesdays.

Only in a leap year this is possible, seven leap years have to go by before this situation can occur again, because in each leap year the 29th would fall on a different weekday.

Seven leap years means 7 x 4 = 28 years to pass after 1984.

Therefore, it will be the year 2012 when a February again has 5 Wednesdays.

Example 9: It was Wednesday on 1 January 2000. What would be the day on 1 January 2001?

Year 2000 is a leap year; hence there are two odd days which will shift the whole calendar of the year 2001 by 2 days.

So, 1 January 2001 will be 2 days ahead means Friday.

Example 10: In an ordinary year which month begin on the same day of the week?

In an ordinary year, Feb has no odd day.

Therefore, Feb and March begin on the same day of the week.

Also, March & November and April & July begin on the same day of the week.

Example 11: What is the maximum sum of the numbers of Saturdays and Sundays in a leap year?

In leap year, 52 weeks and 2 days.

So, from 52 weeks, 52 Saturdays and 52 Sundays.

Remaining 2 days may be Saturday and Sunday.

So totally, 52+52+2=106

Example 12: Today is Monday. After 61 days, it will be:

Each day of the week is repeated after 7 days.

So, after 63 days, it will be Monday.

After 61 days, it will be Saturday.