Aptitude - Calendar - Discussion

The number of days more than the complete weeks are called ODD DAYS....

In 400,1600,2000 odd days are zero...
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Day of the week on 28th may 2006,

In 2006,

Jan 31days+Feb 28days+Mar 31days+Apr 30days+May 28days[bcz 28th may]= 148days [totally]

To find odd day,
148/7=21quotient 1remainder

so 21complete weeks and 1 odd day in 2006

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For 5years [2001-2005],

To find leap year,
5/4=1 [never take the decimal points]
1leap year

So 2odd days in that year[366/7=52weeks+2odd days]
And rest of the 4years have 4odd days[365/7=52weeks+1odd day]

so for 5years,
2+4=6odd days

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Total number of odd days,

0+0+1+6=7

So, 7/7=1complete week
There is zero odd day in it.

0-Sunday
1-Monday
2-Tuesday
3-Wednesday
4-Thursday
5-Friday
6-Saturday


So 28th may 2006 is Sunday....:)

As per the video that I saw on you tube regarding calendar. From freshers world channel. As per the concept/tricks.

1). Take last 2 digits of year
2). Divide this last 2 digits by 4 and consider the quotient.
3). Take the date of the question for which day is to be calculated
4). 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 year code as 2006 falls between the year 2000-2099 it's 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 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 + 00 + 28 + 01 + 06 = 41.

41/7 = 6 (reminder).

Days code is : S-0,M-1,T-2,W-3,T-4,F-5,S-6
The answer is 6 th day of the week is Saturday. On 28 th May 2006 it was Saturday. Not Sunday.

Sorry if anyone has answered already I did not go through the solution.
Here goes the solution,

DAYS:-

0-S, 1-M, 2-T, 3-W, 4-TH, 5-F, 6-SAT

MONTH:-

[J F M][A M Ju][JuL A S][O N D]
[0 3 3] [6 1 4][ 6 2 5][0 3 5] // [ ] Just for simplicity

YEARS;-

1600 to 1699 consider code as 6
1700 to 1799 consider code as 4
1800 to 1899 consider code as 2
1900 to 1999 consider code as 0
2000 to 2099 consider code as 6

FOLLOW THE FOLLOWING STEPS:-

1) Take last two digits of given year: _ _
2) Divide the above taken number by 4 and consider the quotient: _ _
3) Take the date: _ _
4) Take the month: _ _
5) Choose the code of the year as specified above: _
6) Add everything and divide by 7 and consider the remainder the result will be from 0 to 7(any one number).

Match with the day that will be the result.

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.

Solution: JFM AMJ JAS OND
1. Choose appropriate month code: 602 503 514 624
2. Leap years since 1/1/2000 (including the year 2000):
3. Add the month code from step 1, the number of leap years, the day of the month, and the amt of years since 2000. Divide this by 7 and take the remainder.
4. Sun = 1, Mon= 2, T = 3...Fri=6, Sat=0,7

Example: January 1,2050
1. month code 6 2. 50yrs/4=12r2. 12+1 (since Feb 2000 had a leap day) = 13
3. 6 + 13 + 1 + 50 = 70. 70/7=10r0
4. remainder of 0 -> January 1, 2050 will be a Saturday

*If the year you are calculating is a leap year and the date is BEFORE February 29, do add the 1*
Example of this: February 18, 2008
1. month code 2
2. 2 (Feb 2000 & 2004 BUT NOT 2008 yet)
3. 2 + 2 + 18 + 8 = 30. 30/7=4r2
4. remainder 2 -> Monday

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.

In every 400 years, there are 303 non-leap years and 97 leap years. ( 400/4 - 3 = 97 )
How? Every fourth year is a leap year, but century years (i.e, ending with 00) are considered leap years only if they are divisible by 400. So every 100th,200th and 300th years are non- leap years and the 400th year is a leap year.

The Number of odd days in every year= 365%7=1.
The Number of odd days in every leap year= 366%7=2.
The Number of odd days in every 400 years= 303*1 + 97*2 = 497.
But since 497 is a multiple of 7 (i.e, 497%7=0) , odd days is same as 0.

Now, the year 2000 has five such 400 years. So number of odd days in 2000 years= 5 * number of odd days in 400 = 5*0 = 0.

Then follow explanation for the next six years as given in solution.

Hope this helps!

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

--------------------
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

Hello friends, I am Nitin.

For these type of questions I have easy solution. But you have to learn two things.

1. Year code.
2. Month code.

Year code:

1600-1699 = 6.
1700-1799 = 4.
1800-1899 = 2.
1900-1999 = 0.
2000-2099 = 6.

Month code 0 3 3 6 1 4 6 2 5 0 3 5.

J F M A M J J A S O N D.

Let's take same question.

28/05/2006.

Step 1:

Add 1st two digit of date with last two digit of year.

28+06=34.

Step 2:

Divide last two digit of year with 4.

06/4 = 1 (we will take quotient not remainder) then,

Step 3:

Add Step 1 + Step 2.

34+1+year code + month code.

34+1+6+1 = 42.

Step 4:

Divide Step 3 by 7.

= 42/7 = 0 (as remainder) so, 0 means Sunday.

Here is the reason for why 1600 and 400 are 0.

Counting of Odd Days:

1 ordinary year = 365 days = (52 weeks + 1 day. ).
1 ordinary year has 1 odd day.

1 leap year = 366 days = (52 weeks + 2 days).
1 leap year has 2 odd days.

100 years = 76 ordinary years + 24 leap years.

= (76 x 1 + 24 x 2) odd days = 124 odd days.
= (17 weeks + days) 5 odd days.

Number of odd days in 100 years = 5.
Number of odd days in 200 years = (5 x 2) 3 odd days.

Number of odd days in 300 years = (5 x 3) 1 odd day.
Number of odd days in 400 years = (5 x 4 + 1) 0 odd day.

Similarly, each one of 800 years, 1200 years, 1600 years, 2000 years etc. Has 0 odd days.