Aptitude - Calendar - Discussion

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.

Thank you all.

Given :.

Jan 1, 2006 is sunday.

Jan 1, 2010 =?.

Consider completed year before 2010 i.e 2009 (since 2010 was starting).

2009=2000+9.

2009=0+9 (since 400, 800, 1200, 1600. Contains '0' odd days).

9/4=2 (2 leap years are there. Here it is divided by 4 because for every 4 years we get leap year).

Leap year=2*2=4.

Non leap =7*1=7.

Total = 11.

Jan 1st=1 day.

11+1=12.

12/7 = 5.

Friday.

If any year divided by 4 with out reminder its year known as leep year.

E.g: 2008/4 = 502 (no reminder)

And in this year were 29 days in Feb.

I want clear method to calculate odd days.

I want briefly to find odd days.

Pavithra You told us a very good way to solve the calendar question but only few things you left i.e related to leap year, this formula doesn't work for leap year, it works only for non-leap year.

@Dg1234.

You can solve leap year by this formula as told by Pavithra but only few things I have changed. Check out.

Formula For Leap Year:

0 - Apr, jul 0 - Sat
1 - Jan 1 - Sun
2 - May 2 - Mon
3 - Aug 3 - Tue
4 - Feb, mar, nov 4 - Wed
5 - Jun 5 - Thu
6 - Sep, dec 6 - Fri
7 - Sat

Note:- To find the odd days for leap year we have to add 6 only with this formula.

Formula: 6 + Day + Month + (Year-1900) + (year-1900)/4.

Eg. 4-feb-2004.

= 6+4+4+(2004-1900)+(2004-1900)/4.
= 6+8+104+(104/4).
= 118+26.
= 144.

= 144/7.

= Remainder = 4.
= Wednesday.

Codes for the calendar.

DAYCODES.

Sun = 0.
Mon = 1.
Tue = 2.
Wed = 3.
Thurs = 4.
Fri = 5.
Sat = 6.

CENTURY CODE.
1600-1699 = 6.
1700-1799 = 4.
1800-1899 = 2.
1900-1999 = 0.
2000-2099 = 6.
2100-2199 = 4.
-
-
Repeats.

MONTH CODE:.
Jan = 0.
Feb = 3.
Mar = 3.
April = 6.
May = 1.
June = 4.
July = 6.
Aug = 2.
Sept = 5.
Oct = 0.
Nov = 3.
Dec = 5.

A leap year January code=-1 and Feb = 2.

For example:- Find the day of the week on 26-jan-1956.

Solution:-.

Date + last two digits of the year + (last two digits/4) + monthcode + centuarycode.

= 26 + 56 + (56/4) + (-1) + 0 (since, 56/4 remainder is' 0' it is a leap year) (when leap year comes above).
=26 + 56 + 14 - 1 + 0 codes we said that jan=-1).
= 95 (we should divide by 7 because we have to find the day of the week).
= 95/7 (remainder '4').
= 4 (Thursday).

You can also check the calendar on this date.

Month code- 033614625035 or 144025036146?

Please explain it.