In this topic we are going to discuss How to solve Calendar
Problems with ease and in faster way. Most of the students face difficulty in
solving the Calendar Problems. So here we will discuss the important rules and
formulas which will helps in solving the problems.
By using simple mathematics, these formulas can be applied easily. For both
leap and nonleap year we will use different formula with minor changes. Using these formulas all the aspirants can
easily solves the problems in UPSC, SSC, State PSC and other govt job exams.
Leap year rule:
Different types of Calendar Problems:
 Find the day on date 11 Aug, 2017?
 Calendar 2014 will repeat itself in which year?
 If today is Monday then next year on this date, the day will be?
General Concepts:
 a month has 4 weeks
 a month has 30 days
 a year has 52 weeks
 a year has 365 days
Properties
of a year:
 There are two kinds of years: Ordinary year and Leap year.
 An ordinary year has 365 days and a leap year has 366 days, one day extra. This extra day appears in February which in a leap year has 29 days instead of 28.
A year to be
a leap year, it must satisfy two conditions. First, it must be divisible
by 4, such as 2016, 5324 and so on. By this rule every century year should be a
leap year. But for a century year to be a leap year, there is a second
rule, a century year must be divisible by 400 to be a leap
year. 1600, 2000, 2400 are leap years but 1900, 2100, 2200 are not.
An ordinary year has 52 weeks
and 1 odd day, while a leap year has 2 odd days as it has 1
day extra in February.
PROBLEM 1 : FINDING DAY ON A PARTICULAR DATE?
For solving such problems we have to first the year whether it is leap or not. So follow the steps given below.
Lets take an example : Find the day on 9 Feb 1993.
STEP 1: CHECK FOR LEAP YEAR
Find the given year in date is leap year on not. To do this divide the last two digits of given year with 4 .
In above date 09th Feb 1993, divide last two digits with 4 . So, here we get 23 as quotient (forgot the remainder).
In above date 09th Feb 1993, divide last two digits with 4 . So, here we get 23 as quotient (forgot the remainder).
STEP 2: MONTHS CODE
Now we have to remember code 033614625035.
Don't get confused , this is the code for months in a year.
Month

Code

January

0

February

3

March

3

April

6

May

1

June

4

July

6

August

2

September

5

October

0

November

3

December

5

In above date , given month is February. So , according to table code for February is 3.
STEP 3 : CENTURY CODE
Now we have to remember the centuries code .
Century

Code

17001799

4

18001899

2

19001999

0

20002099

6

21002199

4

22002299

2

Now according to date year 1993 lies in century 10991999 i.e code for century is 0.
Step 4 : IMPLEMENTATION IN FORMULA
For NonLeap Year
=(Year Last two digits + Date + Century Code + Month Code + Quotient we got in step 1) ÷ 7
The remainder we get is the day of the week.
According to Date :
= (93+09+0+3+23) ÷ 7
=128÷ 7
= we get 2 as remainder i.e it is the day of the week
Lets check it in table
Lets check it in table
SUNDAY

0

MONDAY

1

TUESDAY

2

WEDNESDAY

3

THURSDAY

4

FRIDAY

5

SATURDAY

6

So according to table we get Tuesday as day on date 09th Feb,1993
For Leap Year :
=(Year Last two digits + Date + Century Code + Month Code + Quotient we got in step 1  1) ÷ 7
For leap year we just have to subtract 1 from the addition in formula.
PROBLEM 2 : CALENDAR REPEATS ITSELF
For Leap Year :
A leap year calendar will repeats itself after 28 years.
For NonLeap Years:
If we start in 2003 then the repeats are in 2014, 2020, 2025 and 2031. The pattern is 11, 6, 5, 6 (starting in 2003).
PROBLEM 3 : DAY IN NEXT YEARS
If it was Sunday on Jan 1, 2006. What was the day of the week Jan 1, 2010?
On 31st December, 2005 it was Saturday.
Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days. Here 2006,2007 and 2009 are non leap years so there is 1 odd day and in leap year means 2008 we add 2 days.
On 31st December 2009, it was Thursday.
Thus, on 1st Jan, 2010 it is Friday.
IF YOU NEED ANY HELP YOU CAN COMMENT BELOW.
Post a Comment