Aptitude - Calendar - Discussion

Why do they take 1600 and 400 every time they solve a problem and in some case 300 years and 1600?

@Masthan.

Please explain this. It's very short. But how to do this?

x+2+1+25+6 = 3.
x+34-3 = 0.

x+31 = 0.
x+28+3 = 0 (28/7 = 0).

x+3 = 0.
4+3 = 7 = 0 (7/7 = 0).

x = 4, 11, 18, 25.

Guys it is easy to find solution to this problem by formula.

Day = Date+Month code+(x-1990)+(x-1990)/4.

Here given day is Wednesday it's code is 4 so now 4 = Date+0+101+25.

4 = Date+126.

You should get (date+126)/7 you should get 4 by trail and error method if you substitute any one day you will know the answer.

Please took 19/08/2015 & explain?

Hi can anyone explain why you are taking 1600 or 300 or 400 at the initial stage please explain.

Hi, I had a little confusion.

Why in some cases it is 1600, 300 years and in some cases it is 1600, 400 years. Kindly explain this.

Why you have not taken 1year in consideration?

Can we use Zellers rule here?

How you found 1600 years and 400 years separately?

19.8.2015 was a wednesday.

Solution:

I. Take the year part and split it into two parts (ie) 2014 & 2015
II. 2014 ...... Can be splited into 2000 years and 14 years.
III. Odd days calculation: since 2000 is a leap century (2000/400 =50) year so 0 odd day.
...>(14 /4= quotient =3 ,remaider=2) gives;

Leap year =3 . Ordinary year =(14-3=11) 11.
.....>Odd days in 14 yrs= [(3*2=6)+(11*1=11)]=6+11=17
Therfore total no of odd days in 2014= 17.
IV. 19.8.2015

Odd days:
Jan Feb Mar Apr May Jun July Aug
31 28 31 30 31 30 31 19
Add all days= 231,
Therfore odd days upto 19.8.2015 = 231,
V.Total odd days = 17(from step III) + 231(from step IV).
=248 /7 = 3(Remainder =3, And quotient =35).

0 -sunday
1- monday
2- tuesday
3- wednesday
4- thursday
5- friday
6- saturday

Therfore answer is Wednesday.

Notes:
Century leap years =400 , 800,1200,1600 ,2000,2400. That is a year which is divisible by 400.
100 is not a leap year eventhough it is divisible 4. Centuary year should be divisible by 400.