Aptitude - Calendar - Discussion

28may2006.
2005
[2000] and[5]
[0] leap year [1]leap year+[4] normal year.
[1*2+4*1] + [3+0+3+2+28].
=42.
42/7.
0 remainder
Sunday Ans.

28TH MAY 2006.
=>1600+400+5 YEARS+[JAN TO APRIL]+28 DAYS [IN GIVEN QUESTION]
=>0+0+[1 LEAP YEAR+ 4 ORDINARY YEAR]+[3+0+2+3]+28.
=>0+[1*2+4*1]+8+28,
=>6+8+28,
=>42 WHICH IS DIVIDED BY 7.
THEN 6 WEEKS + 0 ODD DAYS.

0 = SUNDAY.

Day = date +month code +century code + year + leap year ÷ 7.
Day = 28+2+6+6+1÷7 = 1.
And 1 = Sunday.

Why not consider 2004 is the leap year why are you starting with the 2000 leap year?

Day codes:

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

Month codes:

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

Day codes:
0 1 2 3 4 5 6
S M T W T F S

28 may 2006.
Add 28+06 = 34.

6÷4 =1 with two remainder.

34 + 1 = 35,
35/7=0 remainder.

0 = Sunday.

As per the video that I saw on you tube regarding calendar. From freshers world channel. As per the concept/tricks.

1). Take last 2 digits of year
2). Divide this last 2 digits by 4 and consider the quotient.
3). Take the date of the question for which day is to be calculated
4). Month code of may is 1
[ Similarly as per trick code for various months is : J-0,F-3,M-3,A-6,M-1,J-2,J-6,A-1,S-5,O-0,N-3,D-5]
5). Now take year code as 2006 falls between the year 2000-2099 it's code is 6.
[Similarly for various years range code is:
1600-1699 : 6
1700-1799 : 4
1800-1899 : 2
1900-1999 : 0
2000-2099 : 6]

Now add sum of all these values and divide by 7. As we have to calculate day. Consider the reminder. ie, 06 + 0 + 28 + 1 + 6 means add 06 + 00 + 28 + 01 + 06 = 41.

41/7 = 6 (reminder).

Days code is : S-0,M-1,T-2,W-3,T-4,F-5,S-6
The answer is 6 th day of the week is Saturday. On 28 th May 2006 it was Saturday. Not Sunday.

According to me, the simplest way of solution is;

Months - 1440, 2305, 6146

Here, 1 = January, 4 = February, 4 = March, 0= April, May = 2, June = 3, July = 0, August = 5, September = 6, October = 1, November = 4, December = 6.

Learn this table along with the nos of their respective months.

Now come to days, For Days Remember:
1 - Sunday.
2- Monday.
3- Tuesday.
4- Wednesday.
5 - Thursday.
6- Friday .
0 or no remainder = Saturday.

Now come to century part

Remember the small table and understand the concept as to how we can calculate the no that needs to be assigned to a Century.

The No for 2000 Century must be 6, If we go before we need to again Start the chains or simply started from 6 till you reach 6 no.

For Calculating no, after 2000 century like 2100, 2200 just Substract by 2 until you reach 0

Let's see the below example, to go back century just set the 0,2,4, 6 in increasing order and to go after 2000 century just reduce the no by 2.

In below example 2000 is our base.

1600 - 6
1700 - 4
1800- 2
1900 - 0
2000 - 6
2001 - 4
2002 - 2
2003 - 0

And so on

Now come to the question, it Will hardly take 5 to 10 seconds to solve this qsn now, Just learn the above 3 tables

The question is 28 May 2006

Here write in the following manner

28 - Date.
2 - Month(See no assigned in table),
06 - Year(write last 2 digits of year),
6 - Century(2000),
1 - (Divide last 2 digits by 4).

Add on these nos - 43.

Now Divide 43 by 7 and the remainder would be = 1.

Now see the day's table, where 1 = Sunday.
Hence the answer is Sunday.

HERE AS IT IS GIVEN 28 MAY 2006.


28+ 1(CODE OF MAY) +06(LAST TWO DIGITS OF 2006)+06/4=(1),
SO 28+1+6+1 =36 then divide it by 7 for odd days,
36/7 = 1 is reminder.

So according to days
0 - SUNDAY
1 - MONDAY
2 - TUESDAY
3 - WEDNESDAY
4 - THURSDAY
5 - FRIDAY
6 - SATURDAY
7 - SUNDAY.

Here 1 is answer, So MONDAY (But) as 2006 is a leap year - 1 DAY SO (MONDAY - 1 = SUNDAY).

That's it. Thanks.

For all those who have confusion in this:

We calculate the years from 0 till 2005 so we split 2000 into 400 and 1600 and its odd day is 0, now left from 2000 to 2005 (5 yrs) = 4 ordinary years and 1 leap year = 4*1 + 2*1 = 6 the remainings.

I hope you all get it.

How to find out the odd days? Please explain.