Calender Problem Basics
Calender Problem Basics
Calendar Problem Basics: Problems based on Calendars forms an important part for various competitive exams. This topic involves a lot of logical discussion and analysis.
Calendar: A Calendar is a chart or series of pages showing the days, weeks and months of a particular year, or giving particular seasonal information.
Basic Structure of a Calendar: The calendar shows the different days of a year. As we know a normal year consists of 365 days and a leap year consists of 366 days. Each year is divided into months. The 365 days are divided into 12 months. The months January, March, May, July, August, October and December have 31 days, whereas April, June, September and November have 30 days. The month of February has 28 days in a normal year and 29 days in a leap year. The Leap Year is after every four years. As a Year has 365 days and 6 hours. The 6 hours of these four year are collected to make 24 hours hence one day extra. This extra day is added to the month of February.
Leap Year: A year which is divisible by 4 is a leap year. If it is also divisible by 100, then we have to check it by dividing it by 400.
1988, 2008, 2012, 2016, etc.are all leap years and therefore, are divisible by 4.
2000, 2400,1600, etc. are all leap years and therefore, are divisible by 400 ( both 100 and 4).
1700, 1800, 2100, 2200, etc. are not leap years as they are not divisible by 400 ( even if they are divisible by 4).
Concept Of Odd Days: To find the day of the week on a particular date. We will calculate the days that are extra with respect to a week known as odd Days. We will divide the number of days with 7 and the remainder will be number of odd days. For example if we divide 365 with 7 we get the remainder as 1 . So an ordinary year will have one odd day.
COUNTING OF ODD DAYS
An ordinary year has one odd day, whereas a leap year has two odd days.
A number of odd days in a month
January has 31 days, irrespective of whether it’s an ordinary year or leap year. The division of the number 31 by 7 provides the remainder 3 hence January has 3 odd days. On generalising, any month which has 31 days has 3 odd days and any month which has 30 days has 2 odd days.
The only exception happens is in the case of February. The February month of an ordinary year has 28 days, division of 28 by 7 provides zero as remainder. Hence, the number of odd days in February of an ordinary year will have 0 odd days and that of leap year will have 1 odd day as February in a leap year has 29 days.
The below table depicts the number of odd days in different months of a calendar year:
| Month | Number of odd days |
| January | 3 |
| February(ordinary/leap) | (0/1) |
| March | 3 |
| April | 2 |
| May | 3 |
| June | 2 |
| July | 3 |
| August | 3 |
| September | 2 |
| October | 3 |
| November | 2 |
| December | 3 |
Day of the week repeats itself after every 7 days.
so, if it is a Thursday on 1 Aug, 2013, then it is a Thursday on
Now, we are sometimes confronted by a problem like, if 10 Aug is a saturday, what day it will be on 27 Aug.
FIrst, let us find the number of days from 10 Aug to 27 Aug.
27- 10 = 17 days. ( 10 Aug not counted )
Now, if we divide 17 by 7, it leaves a remainder of 3.
Now, these 3 days can be called odd days, which are left off after a complete week.
As 10 Aug was a saturday, the first odd day would be Sunday.
So, the third odd day is Tuesday . Hence, 27 Aug will be a Tuesday.
Ex. How many odd days are there in a normal year and a leap year?
Sol. A normal year has 365 days and a leap year has 366 days.
So, a normal year would have 1 odd day, and a leap year would have 2 odd days.
Which of the following is a leap year ? | |||
| Right Option : A | |||
| View Explanation | |||
How may odd days are present in a 100 year? | |||
| Right Option : D | |||
| View Explanation | |||
How many odd days are present in March ? | |||
| Right Option : C | |||
| View Explanation | |||
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