1.Odd Days: Number of days more than the complete number of weeks in a given perios is the number of odd days during that period.
2. Leap Year: Every year which is divisible by 4 is called a leap year. Thus, each one of the years 1992, 96, 2004, 2008 etc is a leap year.
Every 4th century is a leap year but no other century is a leap year. Thus each one 400, 800, 1200, 2000 etc are is a leap year. None of 1900, 2010, 2020, 2100 etcis a leap year.
An year which is not a leap year is called an ordinery year.
3. (i) An ordinary year has 365 days
(ii) A leap year has 366 days
4. Counting of Odd days
(i) 1 ordinary year = 365 days= (52weeks + 1day)
An ordinary year has 1 odd day
(ii) 1 Leap year =366 days= (52weeks + 2days)
A leap year has 2 odd days
(iii) 100 years= 76 ordinary years + 24 leap years
= [(76x52)weeks + 76days] + [(24x52)weeks + 48 days)
=5200weeks + 124days
=(5217weeks + 5days)
Therefore 100 years contain 5 odd days
200 years contain 10 and therefore 3 odd days
300 years contain 15 and therefore 1 odd day
400 years contain (20+1) and therefore 0 odd days
Similarly, each one of 800, 1200, 1600, 2000 etc contain 0 odd days
Remark: (7n+m) odd days, where m<7 days=" 1">
5.Counting Days:
We have : Sunday for 0 odd days; Monday for 1 odd day; Tuesday of 2 odd days and so on
Ex.1 Find the day of the week on 16th July, 1776
Sol. (i) 16th july, 1776 means = (1775 years + period from 1st January to 16th July)
Now 1600 years have 0 odd days
100 years have 5 odd days
75 years = (18 leap years + 57 ordinary years)
=(36+57) odd days = 93 odd days
=(13 weeks + 2days) of odd days
= 2 odd days
1775 years have (0+5+2) odd days i.e. 0 odd days
now days from 1st jan, 16th july 1776
Jan Feb March April May June July
31 + 29 + 31 + 30 + 31 + 30 + 16 = 198 days
==(28 weeks + 2 days) = 2 odd days
Number of odd days = (0+2) =2
The days of the week was 'Tuesday'
ex2. what was the day of the week on 12th January, 1979?
Sol. Number of odd days in (1600 + 300) years = 0+1 =1
78 years = (19 leap years + 59 ordinary years)
=(38 + 59) odd days = 97 odd days = 6 odd days
12 days of january have 5 odd days
. . Total number of odd days = (1 + 6 + 5) = 6 odd days
. . The desired day was 'Friday'
Ex.3 Find the day of the week on 25th December, 1995
Sol. 1600 years have 0 odd days
300 years have 1 odd days
94 years = (23 leap years + 71 ordinary years)
=(46 + 71) odd days = 117 odd days = 5 odd days
Number of days from 1st Jan 1995 to 25th Dec 1995 +359 days
= (51 weeks + 2 days) = 2 odd days
. . Total number of odd days = (0 + 1 + 5 + 2) odd days = 1 odd days
Required day is 'Monday'
Ex.4 On what dates of October, 1994 did Monday fall ?
Sol. Find the day on 1st October, 1994
1600 years contain 0 odd day
300 years contain 1 odd day
93 years = (23 leap years + 70 ordinary years)
=(46 + 70) odd days =116 odd days = 4 odd days
Days from 1st Jan, 1994 to 1st Oct 1994
Jan(31) + Feb(28) + March(31) + April(30) + May(31) + June(30) + july(31) + Aug(31) + Sept(30) +Oct(1) = 274 days = 39 weeks + 1 day = 1 odd day
Total Number of odd days = (0+1+4+1) =6
1st Oct, 1994 was 'Saturday'
Monday fell on 3rd October 1994
During October 1994, Monday fell on 3rd, 10th, 17th and 24th
Ex.5. Prove that the calender for 1995 will serve for 2006
Sol. In order that the calender for 1995 and 2006 be the same, 1st january of both the years must be on the same day of the week. For this, the total number of odd days between 31st Dec. 1994 and 31st Dec 3005 must be zero.
We Know that an ordinary year has 1 odd day and a leap year has 2 odd days. During this period there are 3 leap years, namely 1996, 2000 and 2004 and 8 ordinary years.
. . Total number of odd days during this period (6+8)+14=0 i.e. 0 odd day
Hence, the calender for 1995 will serve for 2006
Ex6. Prove that any date any date in march is the same day of the week as the corresponding date in November of the year
Sol. In order to prove the required result, we have to show that the number of odd days between last day of Febraury and last day of October is zero.
Number of days between these dates are:
March(31) + April(30) + May(31) + June(30) + July(31) + Aug(31) + Sep(30) + Oct(31)
=241 days = 35 weeks
Number of odd days during this period = 0
Hence the result follows.
Ex. Prove that the last day of a century cannot be any of Tuesday, Thursday or saturday
Sol. 100 years contain 5 odd days
Last day of 1st century is 'Friday'
200 years contain (5x2)=10 odd days= 3 odd days
Last day of 2nd century is 'Wednesday'
300 years contain (5x3)= 15 odd days = 1 odd day
Last day of 3rd century is 'Monday'
400 years contain 0 odd day
Last day of 4th century is 'Sunday'
Since the order is continually kept in successive cycles. we see that the last day of a century can not be tuesday, Thursday or Saturday
