Aptitude - Calendar - Discussion

@ Sundar.

Thank you for giving a good explanation.

@All.

Here is the clarification.

Most of us have no prob in calculating no. Off odd days.
But confusion is there when considering the day code.

Simple;
1jan 2000 -sundays.
To
1jan 2006.

Odd days = 6.
2006 = 1.
2007 = 1.
2008 = 2.
2009 = 1.

1 jan 2006 so one day =one odd day.

Total=6
Odd day=7-6=1.

Starting day is sunday so again start from sunday then keep increasing by one
For odd 2=
1 = Monday,
2 = Tuesdays .
So on.

Solution:

Day = date+ month code + year code +no.of normal years + no.of leap years-1/7
01 + 01 + 06 + 10 + 10/4 - 4 = 7 - 1 = 6.
6 is Friday,

@All.

Leap year means February has 29 days.
The last leap year was 2020.
Leap year comes every 4 years.
Eg;
2004, 2008, 2012,2016,2020,2024.

Thanks everyone for explaining.

How do we know the leap year of that specific year? How to know? Please explain.

Here Explanations are great. Thanks.

Month code for ordinary year 144 025 036 146.

Formula to find day = date + month code + last 2 digit of year + last 2 digit/4 ......note- take only whole number not after decimal.

Ex.. 15 Aug 1989.

15 + 3 + 89+ (89/4).
15 +3+ 89+22..........89/4=22.xxxx
= 129.

Divide 129 by 7.
129/7 = The remainder will be 3.

Then sunday =1
Monday=2
Tuesday=3
Wednesday=4
Thursday=5
Friday=6
Saturday=00
Aur remainder is 3 then answer will be Tuesday.

Pls, remember.

For 1900 series add in addition = 00
For series 2000 add +6
For series 2100 add +4
For series 2200 add +2
For series 2300 add +00
For series 1800 add +02
For series 1700 add +04
For series 1600 add +06

Hi Friends,



Odd days of a year X = Remainder of [No. of days in the year X / 7].



[Where 7 --> For every 7 days, the day will be repeated, thats why we are dividing the total number of days of an year by 7. If it has any remainder then it will be considered as an odd day for that year.]





The required remainders (odd days) for the different years are given below:



For 2006: 365/7 = 1 (Total no. of days in 2006 is 365)

For 2007: 365/7 = 1

For 2008: 366/7 = 2 (Leap year, hence Feb month has 29 days)

For 2009: 365/7 = 1



Therefore, total number of odd days = 5.



Given day is 'Sunday' and total no of odd days is '5'.



The required date = Sunday + 5



= (The 5^th day next to Sunday)



= Sunday + 1-Monday + 2-Tuesday + 3-Wednesday + 4-Thursday + 5-Friday.



= Friday (Required answer).

If 18th may is a Sunday, then which day will fall on 5th September of the same year?

Can you please find the answer?

I am not getting it.