Calendar problems solving tips without any formula

If you keenly observe the calendar, if today is Wednesday same date of the next year will be Thursday (if there is no Feb 29 in between) and Friday(if Feb 29 comes) but what if someone asks you to find the day of a given random date?

I have found many people asking calendar problems formula. there is no need for any formula as you can do it all by yourself with calendar problem aptitude tricks. 

So here is a simple procedure with which you can simply find the day of a given date.

Before learning the complete procedure, let me introduce a few basic terms required to solve these type of problems.

Calendars are prepared based on the earth's orbit around the sun and it is designed it in a way that every orbit is equivalent to 1 year. with some fraction of the day left out for a year and this fraction is adjusted with the help of leap days.

Most of the people think that every 4th year is a leap year but this is not true if one considers every 4th year as a leap year, for every 400 years they are 3 days ahead of Earth's position w.r.t sun.

Leap Year: 

• A year which is divisible by 400 is a leap year.
• A year which is not divisible by 100 and divisible by 4 is a leap year.

Century: Every
100 years constitutes a century.

E.g 16th century (1501-1600)

E.g 21st century (2001-2100)
Calendar shortcut tricks are derived using the odd-days concept.
Odd days:

Odd days are the extra days remained after splitting the entire days into weeks. For example, 10 days is equivalent to 1 week 3 days so 3 days are considered as odd days in this context. One can simply calculate the remainder when days is divided by 7.

After every 7 days, the cycle repeats. and this the base of the theory odd days.

Instead of counting all the days from a known reference day to the given date we calculate remaining extra days.

For all the problems solved here, reference day considered is "01-01-01" which was Monday.

Calculating odd days:

A leap year has 366 days(52 weeks + 2 odd days) 

A non leap year has 365 days(52 weeks + 1 odd day)

If we calculate for 100 years, we get 24 leap years and 76 non-leap years

Odd days for 100 years is 24*2+76 = 124(17 weeks 5 days) = 5 Odd days

For 200 years(100*2) : 5*2(1 week 3 days) = 3 Odd days

For 300 years(100*3) : 5*3(2 weeks 1 day) = 1 odd day

For 400 years(100*4) : 5*4 and a leap-day(as multiples of 400 are leap years) = 21(3 weeks) = 0 Odd days
at last, Number of odd days in a century = 5.

Day of the Week Related to Odd Days:

No. of days:	0	1	2	3	4	5	6
Day:	Sun.	Mon.	Tues.	Wed.	Thurs.	Fri.	Sat.

Month-wise Odd days:

MONTH      DAYS           ODD DAYS(non-leap/leap years)
Jan              31                 3
Feb              28/29            0/1
Mar              31                 3
Apr              30                  2
May             31                  3
Jun              30                  2
Jul               31                  3
Aug             31                  3
Sep             30                  2
Oct              31                  3
Nov             30                  2
Dec             31                  3
.................................................
total extra days:             29/30
Odd days:                       1(non-leap)/2(leap)

How to remember the number of days in each calendar month?
There is a simple technique with which we can remember the number of days in every calendar month. infact we need nothing to do this, what all you need is just the hands.

This technique can answer the following questions:

• which months have 30 days.
• list of months with 31 days.
• how to remmeber the number of days in any calendar month.

What is this technique?

It is also known as knuckle technique. put your knuckles as shown in the figure and start labeling each up and down as one-month till December and now check which month you are looking for. if it is in the down, it have 30 days(except for february) and if it is up, it have 31 days.

**********************************************************************************************************************************How to solve calendar problems?:

Example problem:

which day was 25th September 96th year of 20 th century?

solution:

20th century begun at 1901 ,96 th year of this century is 1996.
Divide entire days as 
1900 years
+95 years
+8 months 
+25 days

**Part 1:(1-1900)
1900=1600(0 odd days)+300(1 odd days)
Odd days: 1

**Part 2:(1900-95)[we get 23 leap years]
95 years= 23 leap years,72 non leap years.
Extra days: 23*2+72 =46(6 weeks 4 days) + 72(10 weeks 2 days) 
Odd days: 6

**Part 3:(96 JAN-96 SEP):

MONTH      DAYS           ODD DAYS
Jan              31                 3
Feb              29(leap)        1
Mar              31                 3
Apr              30                  2
May             31                  3
Jun              30                  2
Jul               31                  3
Aug             31                  3
Sep             30(25)            4
Oct              31                  0
Nov             30                  0
Dec             31                  0
.................................................
total extra days:              24

odd days:= 3


Final :

Part1 + Part2 + Part3 = 1 + 6 + 3 = 3 odd days.

Hence 25th September was "WEDNESDAY"


** Example 2:
which day was 26 JAN 2015?
given that 25 SEP 1996 was Wednesday
then 25 SEP 2014 would be Thursday(odd days between the two dates =1)
Odd days between 25 sep 2014 and 26 jan 2015=5+3+2+3+26=39(39%7=4)

Thursday+4=Monday

hence 2015 01 26 is MONDAY

Example 3:
Which year calendar can be reused for the year 2011
A)1011   B)1611
C)2111    D)2016
Solution:
Calendar can be reused if odd days between two years is zero and both the years should have same number of days
2011,1611,2111 are non-leap years,
2011-1011=1000 years=3 odd days.
2011-1611=400 years=0 odd days
2111-2011=100 years=5 odd days
2016-2011=5 years
Hence 1611 year calendar can be reused 

You can also do these problems using calendar problems formula.
Important note before winding up the content:*Note: please note that the calendar system before 14 September 1752 was a Julian calendar which considered every 4th year as a leap year. from 14 September 1752, with an adjustment of 11 days Gregorian calendar is proposed.this problem gives a correct value for dates after 1752 (for years before 1752, 11 days adjustment need to be done)

Solving calendar problems is no more a problem.
Still if you want formula for calendar problems, go back, browse it on google and come back as you don't like those techniques.
Hope you have enjoyed the solution,follow us for more things and support us :)

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Tool to check day of a given date

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