Aptitude - Calendar - Discussion

We can calculate the day of week based on current date and day.

Whenever we got back one year on the same date the day of week will be one less than today's day of week.

For example today is 27th january 2014 Monday means 27th january 2013 was one day behind of Monday i.e Sunday.

If the year is leap year the day of week goes back 2 days for examples if 27th january 2013 is Sunday and the previous year 2012 is leap year so 27th january 2012 will be two day behind sunday i.e friday. If like this we calculate we have :

27th january 2013 - Sunday.
2012 - Friday 2011 - thur 2010 - wed 2009 - tue 2008 - sun.
2007 - sat 2006 - fri 2005 - thur 2004 - tue 2003 - mon .
2002 - sun 2001 - sat 2000 - thur 1999 - wed 1998 - tues.

SO now 27th January 1998 was tuesday.
To calculate : 17th june 1998 what day?

Jan Feb Mar Apr May june
5 days 28days 31d 30d 31d 17d Total = 142d
(31-27
including 27)

Now total = 142 days between 27th january 1998(Tuesday) and 17th june 1998 .

To calculate the day divide total no of days by 7 and check the remainder. If remainder is 0 it would be previous day of week. If remainder is 1 it would be same day of week. If remainder 2 it would be next day of week and so on.

So here 142/7 = 20 and remainder 2 so the day of week would be Wednesday(Next day of Tuesday).

There is another way...

Remember the code FOR ALL PROBLEMS ABOUT CALENDER


MONTH : J F M A M J J A S O N D
CODE . : 0 3 3 6 1 4 6 2 5 0 3 5

and

CODE FOR THE YEAR
1700 -- 1799 = 4
1800 -- 1899 = 2
1900 -- 1999 = 0
2000 -- 2099 = 6
2100 -- 2199 = 4

AT FINAL EX: FOR 17 JUNE 1998

take the last two numbers in the year for ex. In 1998 take 98 and divide them by 4 and take only the quotient and leave the reminder

ex: 4)98(24
8
----
18
16
----
2
leave that reminder 2 and take only quotient 24.

add the following

month code for JUNE 4
year code for 1998 0
left quotient while
divided by 4 24
finally add the last
two numbers in year
1998 and date17(98+17) 115

------------
143

divide the amount by 7 i.e., 143/7
here take only the reminder and left the quotient.

ex: 7)143(2
14
----
03
take the value of reminder 3.


WEEK S M T W T F S
CODE 0 1 2 3 4 5 6

IN THE ABOVE PROBLEM ANSWER IS 3 THEN DAY IS W-Wednesday


ANOTHER EXAMPLE: 15 AUGUST 1947

ADD

month code 2
year code 0
quotient divided by 4 11
(47+15) 62
----
75
75/7 take only reminder 5
Answer is 5-F-Friday.

For those who did not understand this problem,let me try to explain you.

The first thing to be cleared is that for every 400 years there are 0 odd days.

This is because,in 400 years,there are 97 leap years and 303 ordinary years(remember that 100,200,300 are not leap years but 400 is a leap year.Because any year ending with 00,that year should be divisible by 400 in order to become a leap year)

So the number of odd days at the end of 400 years will be [(97*2)+(303*1)]=497 days.Convert this into weeks,we get 71weeks + 0odd days=totally 0 odd days

Now coming to the problem,
1997=1600+397
for 1600 years 0 odd days
therefore 397 years can be classified as
1601-1700-->24 leap years + 76 ordinary years=[(24*2)+(76*1)]=
124 odd days
1701-1800-->24 leap years + 76 ordinary years=124 odd days
1801-1900-->24 leap years + 76 ordinary years=124 odd days
1901-1997-->24 leap years + 73 ordinary years=121 odd days
totally we have 124*3 + 121=493 odd days
convert this into weeks=70weeks +3 odd days
Hence at the end of the year 1997,we have 3 odd days

Now coming to the year 1998,
the number of odd days=31+28+31+30+31+17=168 odd days(jan,feb,...jun17th)
Convert this into weeks=24weeks +0 odd days
Hence the total odd days=3+0=3 odd days
therefore 17th june 1998=Wednesday

Think it like this!!
Maybe year 0 A.D. is included as an ordinary year rather than as a leap year.
So,
2000 yrs = 499 leap years. (Year: 0 A.D. is be included.)
= 1501 ordinary years
But, 1997 yrs = 498 leap yrs
= 1499 ordinary years
(Year 1998 is not taken because it is of this year's day of 17th of June that we have to figure out.)

1.) Leap years = 366 days (per year)/ 7 = Remainder is 2. Therefore, 2 odd days for every leap year, i.e. to say,
Total odd days = 2 * 498 = 996 odd days
2.) Ordinary years = 365 days (per year)/7 = Remainder is 1. Therefore, 1 odd day for every ordinary year, i.e. to say,
Total odd days = 1 * 1499 = 1499 odd days.

Therefore, total odd days altogether = 2495 days / 7 = 3 odd days.

April and June, & September and November: 30 days each.
Rest of the month: 31 days.
So, for 17th June = 31+28+31+30+31+17 = 168/7 = 0 odd days.

Total odd days = 3 + 0 = 3 odd days.
If 3 odd days:
Start at Sun: Then, 1st is Mon, 2nd is Tues and 3rd is Wednesday.

To find out any day of any year of date we have to go through a formula.

Always remember Code:

Jan Feb Mar Apr May June July Aug Sep Oct Nov Dec
31 (28/29) 31 30 31 30 31 31 30 31 30 31
3 (0/1) 3 2 3 2 3 3 2 3 2 3

So now today is 17/05/2014 is Saturday.(today's date)
Calculate 17/05/1998.

Now leap year(the year which is divisible by 4 or 400) coming between 1998 and 2014 is 2000,2004,2018,2012 means 4.

And year gap between 1998 to 2014 is 16.

Now add 16 to 4 and then divide with 7(days in a week)
the result gives (16+4)/7= 6

So if today is Saturday then 17/05/1998 will be Saturday - 6 = Sunday.(since we are coming to back year so we should differentiate)
Now calculate days between 17/05/1998 to 17/06/1998 is 31 day.

Now divide 7 in 31, result is 3

So if 17/05/1998 is Sunday then 17/06/1998 will be Sunday + 3= Wednesday.(Since we are going forward year we should add)

@All.

Just remember;

100yrs --> 5 odd
200yrs --> 3 odd
300yrs --> 1odd
400yrs --> 0 odd
Now given 1998 jun 17:
We can write 1600 + 300 + 97 + jan1 to june 17

1600 because it gives odd days as 0 because its a multiple of 400.
300 because you know the odd days is 1 from the above table,
Remaining 97 yrs and jan1 to jun17.

For 97yrs -> 24 leap year and 73 ordinary years.
To calculate the odd days.
For all leap year, multiply by 2 and for an ordinary year with 1.

So,24 * 2 + 73*1 = 121.
Now 121/7 = remainder 2 (odd days)

Therefore;
1600 -> 0 odd days,
300 -> 1 odd days,
97 -> 2 odd days.

Jan 1 to jun17 -> we have 3 (31 days), 1 (30 day), 1 (28 day), and 17 days.

Remember:
31days->3,
30days->2,
29->1.
28->0.

So, we get 3 * 3 + 1 * 2 + 1 * 0 + 17 = 9+2+17 = 28 odd days.

For final step:
0 + 1 + 2 + 28 = 31,
31/7=3 remainder.

Chart:
Sun - 0,
Mon - 1,
Tue - 2,
Wed - 3,
Thu - 4,
Fri - 5,
Sat - 6.

So 3 in the chart is *WEDNESDAY*.

Use doomsday algorithm :

Remember that last day of feb , 4th april, 6th july , 8th august , 10th october and 12th december will have the same day of week which is known as doomsday and we need to calculate it in the following way :

For 1900-1999 always take the anchor day as wednesday.
For 2000-2099 always take base anchor day as tuesday.

So now formula for calculating the doomsday:

Doomsday = anchor(wed for 1900-99 tue for 2000-2099) + Y/12(ignore decimal) + remainder(Y/12) + {remainder(Y/12)/4}.

Take only last two digits of the Year.

Based on the formula we have for 17th june 1998 we have:

Doomsday = wednesday + 98/12 + Remainder(98/12) + Remainder(98/12)/4.

Wednesday + 8 + 2 + 2/4.
Wednesday + 8 + 2 + 0 = wednesday + 10 = saturday.

So from the above list mentioned for the nearest doomsday we can see that 6th june is the nearest doomsday.

So 6th june will have the day Saturday.

17th june - wednesday.

The question is what is day of 17 june 1998?

If the year is 1998 that is mean in 100 year gets total 25 leap years if getting 25 leap in 100 year then why using 24 leap year in above comments.

Calculating leap year.
On every 4 years gets 1 leap year.
1901 to 1904 = 1 leap year.
1905 to 1908 = 1 leap year ....

Calculating odd days
I am calculating odd days from 1 jan to 17 june 1998

in jan = 31 days
feb = 28 days
march = 31 days
april = 30 days
may = 31 days
june = 17 days
------
168 days

Dividing by 7 to 168 days (7 mean 7 days in week).

168/7= 24 = 24 weeks + 0 odd days (there is any more days).

If eg., 170 /7 = 24 weeks + 2 odd days.

Only this not understood.

In above discuss why used 1900, 300, 400.

If the year is 1998.

We should have take in 1998 years.

1901 to 2000 years = total 100 years.

1901 to 1997 years = 97 years.

24 leap year + 52 weeks + 1 odd days.

Hi every one,

Here is the best solution for this for Arun and all those who want to know about concepts of using 1600 and then 300 years.


Basics :
odd days are calculated on weeks. If given 8 days, we can have 7 in a week. so here odd days=1 , where as if we have 13 days then this no will be 5.
for 121 days. 121/7=17weeks +2 days. here 2 days are odd days.

now for a year we have total of 365 day .
for leap year 366.

Step1: every ordinary year(365 days)=1 odd
Step 2: Every leap = 2 odd days.
Step 3: 100 years= 76ordinary + 24 leap year.=76+ 48=124 :: 5 odd
Step 4: year 4th century is leap year. means 100,200,300 years will have 5 odd days and 400th year will have 6 odd days.
thats coz. 400th year=25 leap +75 ordinary years=125:: 6 odd days.

Step 5: for 300 years odd days= 5+5+5=15:: 1 odd
Step 6: and for 400 years odd days =5+5+5+6=21= 0 odd days.

Step 7: Rest is explained clearly.

Friends it's so simple, Remember only code:

Year Code:

1600-1699 = 6.
1700-1799 = 4.
1800-1899 = 2.
1900-1999 = 0.
2000-2099 = 6.

Month Code:

Jan = 0
Feb = 3
Mar = 3
April = 6
May = 1
June = 4
July = 6
Aug = 2
Sep = 5
Oct = 0
Nov = 3
Dec = 5

Day Code:

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

Steps to solve:

What was day on 17/8/1998 ?

Step:1 (add day + last two digit of years).

(17+98=115).

Step:2 (divide last two digit of years by 4).

98/4 == 24.

Step:3 (add code value)

115
+24
+04 --- month code.
+00 --- year code.
-----
143.

Step: 4 (divide result step 3 by 7).

143/7 = 3.

Step:5 (match with day code).

3 === Wednesday.