Aptitude - Calendar - Discussion

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.

Pls tell me ....why did you add 1 in the calculation of odd days in 400 years???

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.

Why, (Number of odd days in 400 years = (5 x 4 + 1) 0 odd day.) in this 1 is added ,

please clarify

To Nitin:

Here we have 97 yrs

97/4=24 leap yrs

97-24=73 ordinary yrs

Thank you Arya. I can understand properly your solution.

There is another way...

Remember the code FOR ALL PROBLEMS ABOUT CALENDER


MONTH : J F M A M J J A S O N D
CODE . : 0 3 3 6 1 4 6 2 5 0 3 5

and

CODE FOR THE YEAR
1700 -- 1799 = 4
1800 -- 1899 = 2
1900 -- 1999 = 0
2000 -- 2099 = 6
2100 -- 2199 = 4

AT FINAL EX: FOR 17 JUNE 1998

take the last two numbers in the year for ex. In 1998 take 98 and divide them by 4 and take only the quotient and leave the reminder

ex: 4)98(24
8
----
18
16
----
2
leave that reminder 2 and take only quotient 24.

add the following

month code for JUNE 4
year code for 1998 0
left quotient while
divided by 4 24
finally add the last
two numbers in year
1998 and date17(98+17) 115

------------
143

divide the amount by 7 i.e., 143/7
here take only the reminder and left the quotient.

ex: 7)143(2
14
----
03
take the value of reminder 3.


WEEK S M T W T F S
CODE 0 1 2 3 4 5 6

IN THE ABOVE PROBLEM ANSWER IS 3 THEN DAY IS W-Wednesday


ANOTHER EXAMPLE: 15 AUGUST 1947

ADD

month code 2
year code 0
quotient divided by 4 11
(47+15) 62
----
75
75/7 take only reminder 5
Answer is 5-F-Friday.

For those who did not understand this problem,let me try to explain you.

The first thing to be cleared is that for every 400 years there are 0 odd days.

This is because,in 400 years,there are 97 leap years and 303 ordinary years(remember that 100,200,300 are not leap years but 400 is a leap year.Because any year ending with 00,that year should be divisible by 400 in order to become a leap year)

So the number of odd days at the end of 400 years will be [(97*2)+(303*1)]=497 days.Convert this into weeks,we get 71weeks + 0odd days=totally 0 odd days

Now coming to the problem,
1997=1600+397
for 1600 years 0 odd days
therefore 397 years can be classified as
1601-1700-->24 leap years + 76 ordinary years=[(24*2)+(76*1)]=
124 odd days
1701-1800-->24 leap years + 76 ordinary years=124 odd days
1801-1900-->24 leap years + 76 ordinary years=124 odd days
1901-1997-->24 leap years + 73 ordinary years=121 odd days
totally we have 124*3 + 121=493 odd days
convert this into weeks=70weeks +3 odd days
Hence at the end of the year 1997,we have 3 odd days

Now coming to the year 1998,
the number of odd days=31+28+31+30+31+17=168 odd days(jan,feb,...jun17th)
Convert this into weeks=24weeks +0 odd days
Hence the total odd days=3+0=3 odd days
therefore 17th june 1998=Wednesday

How you got ordinary days? could you explain?

Here is complete solution, the year is 1998
16+3=19
See from 0 A.D
1600yr+300yr =1900yr
We are calling the odd days from the 0 A.D. So 1600 yr has 0 odd days
But the remaing 300 year has 15 odd days and rest you can do.

How to calculate odd days for 300 years? I don't understand this only.