Suppose,
you are asked to find the day of the week on 30^{th} June, 1974, it
would be a tough job to find it if you do not know the method. The method of
finding the day of the week lies in the number of "odd days".

Note : Every 7^{th} day will be the same day
count wise, i.e., if today is Monday, then the 7^{th} day .

counting from Tuesday onwards will once again be Monday. Hence, by dividing the total .

numbers of days by 7, the remainder obtained will be called the odd days.

Example : 52 days ÷ 7 = 3 odd days.

A Non – leap year has 365 days whereas a leap year has one extra day because of 29 days in

the month of February. Every year which is divisible by 4 is called a leap year. Leap year .

consists of 366 days, (52 complete weeks + 2 days), the extra two days are the odd days. So, a .

leap year has two odd days because 366 ÷ 7 = 2 (remainder).

An ordinary year consists of 365 days (52 complete weeks + 1 day), the extra one day is the

odd day. So, an ordinary year has one odd day.

Note : Every century, which is a multiple of 400, is a leap year.

Example : 400, 800, 1200, 1600, .......... are leap years.

100
years consist of 24 leap years + 76 ordinary years. (100 years when divided by
4, we get 25 leap years but 25^{th} i.e. the 100^{th} year is
not a leap year, hence only 24 leap years).

= 2 x 24 odd days + 1 x 76 odd days

= 124 days = 17 weeks + 5 days

The extra 5 days are the number of odd days.

So, 100 years contain 5 odd days.

Similarly, for 200 years we have 10 extra days (1 week + 3 days).

200 years contains 3 odd days.

Similarly, 300 years contain 1 odd day and 400 years contain 0 odd days.

Counting of number of odd days, when only one date is given :

Here, we take January 1^{st} 1 AD as the earlier
date and we assume that this day is a Monday. .

After this, the above mentioned method is applied to count the number of odd days and find the day of the week for the given date.

Counting number of odd days, when two dates are given :

Any month which has 31 days, has 3 odd days.

(31 ÷ 7 leaves 3 as remainder) and any month having 30 days has 2 odd days (30 ÷ 7 leaves 2 as remainder).

Then, the total number of odd days are calculated by adding the odd days for each month. The final figure is again divided by 7 to get the final odd days. Finally, the day of the week for the second date is obtained by adding the odd days to the day of the week for earlier date.

If
you were born on 28^{th} January 1988, which was a Sunday, on what day
of the week will your birthday fall in 1989?

(a) Monday

(b) Tuesday

(c) Sunday

(d) Cannot be determined

Since,
1988 is a leap year and as your birthday is before February 29^{th},
your birthday in the next year will be two days after Sunday (since a leap year
will have two odd days), which is Tuesday.Choice (b).

If 8^{th}
March, 1988, which is your date of birth, is a Monday, on what day of the week
will your birthday fall in the year 1989?

(a) Tuesday

(b) Sunday

(c) Monday

(d) Friday

Since,
8^{th} March, 1988 comes after February 29^{th} the number of
odd days between 8^{th} March 1988 and 8^{th }March 1989 is
only one. Hence, in 1989 your birthday will be one day after Monday, i.e., on
Tuesday.

Choice (a)

If 25^{th} May, 2003 is a Sunday, what day of the
week will be 25^{th} December in that year?

(a) Monday

(b) Tuesday

(c) Wednesday

(d) Thursday

Month : |
May |
+ |
June |
+ |
July |
+ |
Aug |
+ |
Sep |
+ |
Oct |
+ |
Nov |
+ |
Dec |

Number of days : |
6 |
+ |
30 |
+ |
31 |
+ |
31 |
+ |
30 |
+ |
31 |
+ |
30 |
+ |
25 |

Total number of days = 214

Number of odd days in 214 days

= 214 ÷ 7 = 30 complete weeks 4 – 4 odd days.

25^{th} December will be 4 days after Sunday,
i.e., on Thursday.

(or)

Month : |
May |
+ |
June |
+ |
July |
+ |
Aug |
+ |
Sep |
+ |
Oct |
+ |
Nov |
+ |
Dec |

Number of odd days : |
6 |
+ |
2 |
+ |
3 |
+ |
3 |
+ |
2 |
+ |
3 |
+ |
2 |
+ |
4 |

Total number of odd days = 25 and 25/7 = 4 odd days i.e., 4 days after Sunday, i.e., Thursday.

Choice (d)

What day of the week is 8^{th} March, 1990?

(a) Tuesday

(b) Thursday

(c) Friday

(d) Sunday

8th March 1990 = (1600 + 300 + 89) years + Ist January
1990 to 8^{th} March 1990.

1600 years will have '0' odd days.

300 years will have ' 1' odd day.

89 years consists of 22 leap years and 67 non – leap years.

1 leap year will have 2 odd days and 1 non leap year will have 1 odd day.

Number of odd days in 89 years

22 x 2 + 67 x 1 = 111 odd days

Number of odd days from 1^{st} January 1990 to 8^{th}
March 1990.

Month : Jan + Feb + Mar

Odd days : 3 + 0 + 1 = 4

Total number of odd days

1 + 111 + 4 = 116odddays.

116 ÷ 7 = 16 weeks + 4 odd days

4 days means Thursday.

Hence, 8th March 1990 is on Thursday.Choice (b).

Which year will have the same calendar as that of 2001?

(a) 2005

(b) 2006

(c) 2007

(d) 2009

Counting the number of odd days from 2001 onwards.

Year : 2001 + 2002 + 2003 + 2004 + 2005 + 2006

Number of odd days : 1 + 1 + 1 + 2 + 1 + 1

Total number of odd days = 7.

Hence, 2007 will have the same calendar as that of 2001.

Choice (c)