Aptitude - Calendar - Discussion

How did 1600 and 400 come please explain me?

I will give you one example:

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First of all you have to learn imp things...ie;

Step 1:
Odd days in 100 years is 5
Odd days in 200 years is 3
Odd days in 300 years is 1
Odd days in 400 years is 0

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Ok.

Step 2:

1 for Mon , 2 for Tue ..... 7 for sunday.

Step 3:

Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
31 28 31 30 31 30 31 31 30 31 30 31

Step 4:

Example is

Find a day ?

From "28 May 2006"

2005 years + (01-01-2006 to 28-05-06)

CONVERT 2005 as

2005 is same as 1600+400+5

Odd days in 1600 years is 0
Odd days in 400 years is 0

Calculate 5years?

5years = (5/4)leap year + (5- (5/4)) ordinary day
= 1 leap year + 4 ordinary days
= (1*2) ordinary days + 4 ordinary days
= 6 odd days.

Next Step:

Jan+Feb+Mar+Apr+May=31+28+31+30+28=148 odd days.

Convert 148 as (weeks+days)format.

How?

148/7 = (21 weeks)+(1 odd day)
= 1 odd day.

Final Step:

1600+400+5+months+day = 0+0+6+1 = 7.

So DAY is SUNDAY.

Odd days means when you divide by 7 to get the no of weeks the remainder left is known to be odd days.

I solved this problem this way.

step 1 : 28 may 2006: take date and year's last two digit number.

28+06 = 34.

step 2: year is divided by 4 so 06/4=1(leave the remainder).
step 3: Codes.

Jan : 0
Feb : 3
Mar : 3
aprl: 6
May : 1
Jun : 2
July; 6
aug : 2
Sep : 5
oct : 0
Nov : 3
Dec : 5

Now add step 1,step 2,step 3 ie,. 34+1+1=36.

divide it by seven 7, so remainder is 0.

Then,

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

Important Points:

An ordinary year has 365 days = 52 weeks and 1 odd day.

A leap year has 366 days = 52 weeks and 2 odd days.

Century = 76 Ordinary years + 24 Leap years.

Century contain 5 odd days.

200 years contain 3 odd days.

300 years contain 1 odd day.

400 years contain 0 odd days.

Last day of a century cannot be Tuesday, Thursday or Saturday.

First day of a century must be Monday, Tuesday, Thursday or Saturday.


Explanation:

100 years = 76 ordinary years + 24 leap years

= 76 odd days + 24 x 2 odd days

= 124 odd days = 17 weeks + 5 days

100 years contain 5 odd days.


No. of odd days in first century = 5

Last day of first century is Friday.


No. of odd days in two centuries = 3

Wednesday is the last day.


No. of odd days in three centuries = 1

Monday is the last day.


No. of odd days in four centuries = 0

Sunday is the last day.


Since the order is continually kept in successive cycles, the last day of a century cannot be Tuesday, Thursday or Saturday.

So, the last day of a century should be Sunday, Monday, Wednesday or Friday.

Therefore, the first day of a century must be Monday, Tuesday, Thursday or Saturday.


Working Rules:

Working rule to find the day of the week on a particular date when reference day is given:

Step 1: Find the net number of odd days for the period between the reference date and the given date (exclude the reference day but count the given date for counting the number of net odd days).

Step 2: The day of the week on the particular date is equal to the number of net odd days ahead of the reference day (if the reference day was before this date) but behind the reference day (if this date was behind the reference day).

Working rule to find the day of the week on a particular date when no reference day is given


Step 1: Count the net number of odd days on the given date

Step 2: Write:

For 0 odd days " Sunday

For 1 odd day " Monday

For 2 odd days " Tuesday

. . . .

. . . .

. . . .

For 6 odd days " Saturday


Examples:

1. If 11th January 1997 was a Sunday then what day of the week was on 10th January 2000?

Sol: Total number of days between 11th January 1997 and 10th January 2000

= (365 " 11) in 1997 + 365 in 1998 + 365 in 1999 + 10 days in 2000

= (50 weeks + 4 odd days) + (52 weeks + 1 odd day) +

(52 weeks + 1 odd day) + (1 week + 3 odd days)

Total number of odd days = 4 + 1 + 1 + 3 = 9 days = 1 week + 2 days

Hence, 10th January, 2000 would be 2 days ahead of Sunday i.e. it was on Tuesday.


2. What day of the week was on 10th June 2008?

Sol: 10th June 2008 = 2007 years + First 5 months up to May 2008 + 10 days of June

2000 years have 0 odd days.

Remaining 7 years has 1 leap year and 6 ordinary years2 + 6 = 8 odd days

So, 2007 years have 8 odd days.

No. of odd days from 1st January 2008 to 31st May 2008 = 3+1+3+2+3 = 12

10 days of June has 3 odd days.

Total number of odd days = 8+12+3 = 23


23 odd days = 3 weeks + 2 odd days.

Hence, 10th June, 2008 was Tuesday.

Odd days= (that particular year) /7, the remainder that you get is the no.of odd days in it.

Too many different month codes used to do the calculation in the above answer figures !

Guyz we always calculate from 31st DEC 200.. to 31 st DEC 200..

So here, 31 st DEC 2000 to 31st DEC 2005 we gonna calculate

Now year 2001, odd days => 1
2002 => 1
2003 => 1
2004 => 2
2005 => 1

6 odd days and rest of the calculation you can see above.

The fast and easy solution according to me is like this:

Days code:- S=1,M=2,T=3,W=4,TH=5,F=6,S=7/0
Month code
JFM AMJ JAS OND
144 025 036 146

CENTURY CODE
100/ 200/ 300/ 400
500 /600/700/ 800

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4 / 2 / 0 / 6

Ex. 15/8/1947 = day ?
Sol: date+month code+ century code + leap year days + normal year days= odd days/7
15+3 + 0 +11*2 + 36 = 76/7, odd days = 6 by matching odd days, it is FRIDAY

Deena your wrong by the way gave some idea to do calender sum like this thank you.

Naveen your procedure is absolutely prefect and thanks boss.