Aptitude - Calendar - Discussion

We have an easy method.

Follow this if you feel it simple.

This is a easy code technique.

Year cheat code, Month cheat code, and date cheat code are the three types.

YEAR CHEAT CODE:

1600 to 1699 = 6.
1700 to 1799 = 4.
1800 to 1899 = 2.
1900 to 1999 = 0.
2000 to 2010 = 6.

MONTH CHEAT CODE:

Jan = 0.
Feb = 3.
Mar = 3.
Apr = 6.
May = 1.
Jun = 4.
Jul = 6.
Aug = 2.
Sep = 5.
Oct = 0.
Nov = 3.
Dec = 5.

DAY CHEAT CODE:

Sunday = 0.
Monday = 1.
Tuesday = 2.
Wednesday = 3.
Thursday = 4.
Friday = 5.
Saturday = 6.

Now, let's take an example,

11/03/1997 which is my birthday.

Step 1: Take the last two digits of the year = 97.

Step 2: Take the date = 11.

Step 3: Divide 97 by 4 and take the quotient = 24.

Step 4: Take the year cheat code = 0 (For the range 1900 to 1997).

Step 5: Take the Month cheat code = 03.

Now add all the results.

97+11+24+00+03 = 135.

Divide the answer by 7 and consider the remainder = 2 (Subtract 1 from this in case of a leap year).

Which is the Cheat code for Tuesday, therefore the answer is Tuesday.

We have an easy method......
Follow this if you feel it simple :)
This is a easy code technique.

Year cheat code, Month cheat code and date cheat code are the three types.
YEAR CHEAT CODE
1600 to 1699 = 6
1700 to 1799 = 4
1800 to 1899 = 2
1900 to 1999 = 0
2000 to 2010 = 6
MONTH CHEAT CODE
Jan = 0
Feb = 3
Mar = 3
Apr = 6
May = 1
Jun = 4
Jul = 6
Aug = 2
Sep = 5
Oct = 0
Nov = 3
Dec = 5
DAY CHEAT CODE
Sunday = 0
Monday = 1
Tuesday = 2
Wednesday = 3
Thursday = 4
Friday = 5
Saturday = 6
Now lets take an example....
11/03/1997 which is my birthday
Step 1 : Take last two digits of the year = 97
Step 2 : Take the date = 11
Step 3 : Divide 97 by 4 and take the quotient = 24
Step 4 : Take the year cheat code = 0 (For the range 1900 to 1997)
Step 5 : Take the Month cheat code = 03
Now add all the results
97+11+24+00+03 = 135.

Divide the answer by 7 and consider the remainder = 2 (Subract 1 from this in case of leap year).

Which is the Cheat code for Tuesday, therefore the answer is Tuesday.

Year Code Month Code Day Code

1600-1699=6 Jan-0 Sun-0
1700-1799=4 Feb-3 Mon-1
1800-1899=2 Mar-3 Tue-2
1900-1999=0 Apr-6 Wed-3
2000-2099=6 May-1 Thu-4
2100-2199=4 Jun-4 Fri-5
July-6 Sat-6
Aug-2
Sept-5
Oct-0
Nov-3
Dec-5

Note: Day+Year+Leap Year+Year Code+Month Code

First we need to calculate leap year; it is calculated on 2010 and we need to take last two digits i.e;(10).

2)10(5
10
----
0 leave 0 and take only quotient i.e;5

Now come to formula
D+Y+L.Y+Y.C+M.C ==> 15+10+5+0+2 = 32.

2)32(16
2
------
12
12
------
0

The 0 shows day's code.
.'.2010 Aug 15th is Sunday.

Note: While calculating leap year, take last two digits from the year.
And considering year in the problem we need to take last two digits (i.e;10 in the problem).

Thank You & Regards,
Arjun.

friends, Let me introduce a technique which is used to find any day with any date from 1900 to 2999.

This is called doomsday technique;

doomsday means:

Last day of feb.

4th april(04.04)
6th june (06.06)
8th august(08.08)
10th oct(10.10)
12th dec(12.12)

To find doomsday of any year we should do,

anchor+(y/12)+remainder(y/12)+(remainder(y/12)/4).

For any period from 1900 to 1999 anchor = 3(wednesday).
For any period from 2000 to 2999 anchor = 2(tuesday) .
y means last 2 digits of the given year.

Now, our question here is to find the day on 15.08.2010
So,anchor=2, y=10.

2+(10/12)+rem(10/12)+(rem(10/12)/4).
2+0+10+2 =14.
14/7 = 0 so 0 odd days.
So dooms day is sunday.

For our question aug 8 is nearer. So, 08.08.2010 is surely sunday
add another 7 to 0 odd days. because 8+7=15. i.e 8th aug + 7 days = 15th aug.

0+7 odd days=7 odd days=7/7=0.

So the day is Sunday.

The total odd days in 1600 years = 0
similarly total odd days in 400 years = 0.

So total odd days in 2000 years = 0,
after 2000 we have completed 9 years.

In those 9 years, in those 9 years, we have 2 leap years and 7 ordinary years
In a leap year, we have 366 days i.e. 52 weeks and 2 days = 2 odd days
In an ordinary year, we have 365 days i.e. 52 weeks and 1 day = 1 odd day.

So in 9 years = 2*2 + 7*1 = 11 in this 11 we have 1 week + 4 days = 4 odd days.

After those 9 years, the 10th year starts in that year we have Jan, Feb, Mar, Apr, May, Jun, Jul and 15 days of Aug.
So total days = 31+28+31+30+31+30+31+15=227 days = 32 weeks + 3 days=3 odd days.
So total odd days = 3+4=7 odd days = 1 week = 0 odd days.
0=7=Sun,
1=Mon,
2=Tue,
3=Wen,
4=Thu,
5=Fri,
6=Sat.
So, the given day is Sunday.

Hey, this type of questions can be easily done by following formula
---------------------------------------------------------------------------------
f= K+[(13m-1)/5]+ d+[d/4]+[c/4]-2c

Here k= date of which you have to calculate(15 in above que.)
m= month (for march, m=1 and so on..for jan m=11)
d= last two digit of given year(10 in above que.)
c= first two digit of given year(20 in above que.)
---------------------------------------------------------------------------------
note: take only integer value after division for ex. 13/5=2
---------------------------------------------------------------------------------

For given question

15th August, 2010?

f=15+[(13*6-1)/5]+10+[10/4]+[20/4]-2*20
=15+15+ 10 + 2 + 5 -40
=7

Now 7%7==0

So its SUNDAY.

August 15th,2010 = (2000) +
(from 2001 to 2009) +
(from 1st January 2010 to 15th august 2010).

Odd Days in 2000 = 0.
Odd Days (from 2001 to 2009) = 2 Leap Year+7 Ordinary Year
= (2*2)+(7*1)
= 11%7 = 4 Odd Days.

Odd Days (from 1st January 2010 to 15th august 2010):

Jan = 31.
Feb = 28(not a leap year).
mar = 31.
Apr = 30.
May = 31.
Jun = 30.
Jul = 31.
Aug = 15.
===========
Tot = 227.
===========

Find Odd Days : 227%7 = 3(Odd Days).

Answer : Add all Odd Days = 0+4+3 = 7.

Notes for Odd Days : Number of days more than a complete week is
called as "Odd Days".

So, you have to divide 7 by 7.

Final Answer = 7%7 = 0 (Sunday)

15 th August 2010.

Jan 31 + feb 28 + Mar 31 + Apr 30 + May 31 + June 30 + July 31+ Aug 15.

31 + 28 + 31 + 30 + 31 + 30 + 31 + 15 = 227.

227/7 = 3 odd days.

*************************************.

For 10 years.

2000 + 2001 + 2002 + 2003 + 2004 + 2005 + 2006 + 2007 + 2008 + 2009.

(In 400, 1600, 2000 odd days are zero. ).

0 + 1 + 1 + 1 + 2 + 1 + 1 + 1 + 2 + 1 (odd days).

= 11 odd days.

Total odd days = 3 + 11 = 14.

14/7= 7.

We have 2 leap year + 7 ordinary years so =2 * 2 + 7 = 4 + 7 = 11 = 4 odd days
So, 4 + 3 = 7.

7/7 = 0 = Sunday.

@Madhu.

For 15 years we have 3 leap year (12/4=3).
And 12 ordinary years,
So total num of odd days=(3*2+12*1)=18,

(Add 2 point for leap year because 2 days will shift for the next year odd days is the actual concept that if 2016 Jan 1 is sunday then 2017 Jan would be Monday but after a leap year it would be Tuesday so 2 days will shift).

And try to divide the number of odd days with 7.

Here it is 18 cant divide by 7.
So 14/7=2.
Then 4 days are remaining then 4 is the num of odd days.
*(dividing 7 means a week have seven days).

So each year has one odd day and leap year has 2 odd days for every 100 years 100+24 odd days but after every 7 days the same day repeats for example today is Saturday after 7 days will be Saturday basing on this we neglect multiples of 7 and by dividing 124 with 7 we get 5 odd days.

So 100 years have 5 odd days but for every 400 years one day is added based on earth's rotation don't ask me about that so 400 years have 20+1 odd days which means 3 weeks and can be neglected as same day appears after any number of complete weeks.