Aptitude - Calendar - Discussion

According to me, the solution is;

1). Take the last 2 digits of the year
2). Divide these last 2 digits by 4 and consider the quotient.
3). Take the date of the question for which day is to be calculated
4). The 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 the year code as 2006 falls between the years 2000-2099 and its 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 the 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 + 01 + 28 + 01 + 06 = 41.
42/7 = 0 (reminder).

Days code is :
S-0,
M-1,
T-2,
W-3,
T-4,
F-5,
S-6.
The answer is Sunday.

A year in is divisible by 4 and for century years it should be divisible by 400, not 100 then it is said to be a leap year.

For example, 100 is not a leap year even though it is divisible by 4 but as it is a century year it should be divisible by 400.

* Odd days need to be divided the number of days by week -for the normal year 365 days (365/7) reminder — 1 is odd days.
* Similarly for leap years -366 days (366/7) reminder- 2 is odd days.

For 100 years (76 normal years (1 odd day)+ 24 leap years (2 odd days).
76x1 + 24x2= 124 days.
124 days/7 days = 5 odd days.

0 + 2 + 6 + 1 = 9,
9/7 reminder is 2,
2 + 6 = 8.
8/7 reminder 1.
So, it is Sunday = answer.

So,we have to calculate the day that falls on 28th May 2006.

At first we have to find the odd days.

1)28
To find the odd days for 28 divide by 7
We get 0 odd days.

2)May
To find the odd days for may ,we should know the odd days of the months.
Jan- 31 days divide it by 7 we get the remainder 3 so, for Jan we have 3 odd days.
Similarly,

Feb-0 for ordinary years and 1 for leap years.
Mar-3
Apr-2
May-3
June-2
July-3
Aug-3
Sept-2
Oct-3
Nov-2
Dec-3
To find the odd days for may we add the odd days up to Apr.
Then 3+0(as 2006 is not a leap year)+3+2=8 divide it by 7 we get a remainder 1.

3) 2006
Take 2005, nearest leap century year is 2000+5 for 2000 0 odd days, and for 5 years there is a leap year.
1×2 = 2
4×1 = 4
Total = 6 odd days
Add all odd days.
0 + 1 + 6 = 7 ÷ 7 = 0.
Therefore the answer is SUNDAY.